# Direct Products and Quotient Groups

#### Peter

##### Well-known member
MHB Site Helper
In Beachy and Blair: Abstract Algebra, Section 3.8 Cosets, Normal Groups and Factor Groups, Exercise 17 reads as follows:

----------------------------------------------------------------------------------------------------------------------

17. Compute the factor group [TEX] ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2) [/TEX]

----------------------------------------------------------------------------------------------------------------------

Since I did not know the meaning of "Compute the factor group" I proceeded to try to list them members of [TEX] ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2) [/TEX] but had some difficulties, when I realised that I was unsure of whether the group [TEX] ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) [/TEX] was a group under multiplication or addition. SO essentially I did not know how to carry out group operations in [TEX] ( \mathbb{Z}_6 \times \mathbb{Z}_4 ) / (2,2) [/TEX].

Reading Beachy and Blair, Chapter 3 Groups, page 118 (see attachment) we find the following definition:

-------------------------------------------------------------------------------------------------------------

3,3,3 Definition. Let [TEX] G_1 [/TEX] and [TEX] G_2 [/TEX] be groups. The set of all ordered pairs [TEX] (x_1, x_2) [/TEX] such that [TEX] x_1 \in G_1 [/TEX] and [TEX] x_2 \in G_2 [/TEX] is called the direct product of [TEX] G_1 [/TEX] and [TEX] G_2 [/TEX], denoted by [TEX] G_1 \times G_2 [/TEX].

----------------------------------------------------------------------------------------------------------------

Then Proposition 3,3,4 reads as follows:

-----------------------------------------------------------------------------------------------------------------

3,3,4 Proposition. Let [TEX] G_1 [/TEX] and [TEX] G_2 [/TEX] be groups.
(a) The direct product [TEX] G_1 \times G_2 [/TEX] is a group under the operation defined for all [TEX] (a_1, a_2) , (b_1, b_2) \in G_1 \times G_2 [/TEX] by

[TEX] (a_1, a_2) (b_1, b_2) = (a_1b_1, a_2b_2 ) [/TEX].

(b) etc etc

------------------------------------------------------------------------------------------------------------------

However in Example 3.3.3 on page 119 we find the group [TEX] ( \mathbb{Z}_2 \times \mathbb{Z}_2 ) [/TEX] dealt with as having addition as its operation.

My question is - what is the convention on direct products of [TEX] ( \mathbb{Z}_n \times \mathbb{Z}_m ) [/TEX] - does one use addition or multiplication?

Presumably, since the operations involve integers the matter is more than one of notation?

Can someone please clarify this matter?

Would appreciate some help.

Peter

#### mathbalarka

##### Well-known member
MHB Math Helper
Peter said:
My question is - what is the convention on direct products of $\mathbb{Z}_n×\mathbb{Z}_m$ - does one use addition or multiplication?
Since direct product are taken pointwise, the operation is a matter of pairing. If their are two pairs $(a_1, b_1)$ and $(a_2, b_2)$ where the elements of the tuples are as in proposition 2, then direct product results $(a_1 *_{G_1} b_1, a_2 *_{G_2} b_2)$. In the case of two cyclic groups of order n and m respectively, we have one operation mod n and other mod m

#### Deveno

##### Well-known member
MHB Math Scholar
Um...$\Bbb Z_n$ is only a group under addition modulo $n$, since the coset containing 0 never has a multiplicative inverse (one usually does not speak of the trivial group as "$\Bbb Z_0$").

To get a handle on what this quotient group might look like, it is helpful to know the ORDER of (2,2) in this group. It turns out that $\Bbb Z_6 \times \Bbb Z_4$ is of order 24, so we need only consider divisors of 24. In fact, it can be shown that the maximal possible order is lcm(6,4) = 12. We have:

$(2,2) \neq (0,0)$, so (2,2) is not of order 1.

$(2,2) + (2,2) = (4,0) \neq (0,0)$, so (2,2) is not of order 2.

$3(2,2) = (0,2) \neq (0,0)$, so (2,2) is not of order 3.

$4(2,2) = (2,0) \neq (0,0)$, so (2,2) is not of order 4.

5 does not divide 24.

$6(2,2) = (0,0)$, thus (2,2) is of order 6.

Thus our quotient group has order 24/6 = 4.

To avoid awkward notation, lets denote the subgroup generated by (2,2) as $H$. The identity of $\Bbb Z_6 \times \Bbb Z_4/H$ is, of course:

$H = \{(0,0),(2,2),(4,0),(0,2),(2,0),(4,2)\}$

Since our quotient group has order 4, we might hope that an element $a$ of order 4 in $\Bbb Z_6 \times \Bbb Z_4$ generates the cosets. Let's see if this is so:

Choose $a = (0,1)$, which is clearly of order 4 in our direct product group. Since $a \not\in H$, clearly, $a + H$ is a distinct element of $\Bbb Z_6 \times \Bbb Z_4/H$ from $H$. So we have as our second element of the quotient group:

$(0,1) + H = \{(0,1),(2,3),(4,1),(0,3),(2,1),(4,3)\}$.

Unfortunately, we see that $2a = 2(0,1) = (0,2) \in H$, which means that $(0,1) + H$ has order 2 in the quotient group.

However, clearly $(1,0) \not\in H$ nor $(0,1) + H$, so $(1,0) + H$ is a 3rd element of the quotient group:

$(1,0) + H = \{(1,0),(3,2),(5,0),(1,2),(3,0),(5,2)\}$.

Again, $2(1,0) = (2,0) \in H$, so $(1,0) + H$ is of order 2 in the quotient group. This tells us the quotient group is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, since it has more than one element of order 2.

For completeness' sake, we list the last element of the quotient group (which is the set of all the elements of the original direct product not yet listed in any coset) namely:

$(1,1) + H = \{(1,1),(3,3),(5,1),(1,3),(3,1),(5,3)\}$

It should be clear that the mapping $\phi:\Bbb Z_2 \times \Bbb Z_2 \to \Bbb Z_6 \times \Bbb Z_4/H$ given by:

$\phi(a,b) = (a,b) + H$

is an isomorphism.

Looking at it the other way, you should be able to come up with a surjective homomorphism from $\Bbb Z_6 \times \Bbb Z_4 \to \Bbb Z_2 \times \Bbb Z_2$ with kernel $H$. Can you?

#### Peter

##### Well-known member
MHB Site Helper
Um...$\Bbb Z_n$ is only a group under addition modulo $n$, since the coset containing 0 never has a multiplicative inverse (one usually does not speak of the trivial group as "$\Bbb Z_0$").

To get a handle on what this quotient group might look like, it is helpful to know the ORDER of (2,2) in this group. It turns out that $\Bbb Z_6 \times \Bbb Z_4$ is of order 24, so we need only consider divisors of 24. In fact, it can be shown that the maximal possible order is lcm(6,4) = 12. We have:

$(2,2) \neq (0,0)$, so (2,2) is not of order 1.

$(2,2) + (2,2) = (4,0) \neq (0,0)$, so (2,2) is not of order 2.

$3(2,2) = (0,2) \neq (0,0)$, so (2,2) is not of order 3.

$4(2,2) = (2,0) \neq (0,0)$, so (2,2) is not of order 4.

5 does not divide 24.

$6(2,2) = (0,0)$, thus (2,2) is of order 6.

Thus our quotient group has order 24/6 = 4.

To avoid awkward notation, lets denote the subgroup generated by (2,2) as $H$. The identity of $\Bbb Z_6 \times \Bbb Z_4/H$ is, of course:

$H = \{(0,0),(2,2),(4,0),(0,2),(2,0),(4,2)\}$

Since our quotient group has order 4, we might hope that an element $a$ of order 4 in $\Bbb Z_6 \times \Bbb Z_4$ generates the cosets. Let's see if this is so:

Choose $a = (0,1)$, which is clearly of order 4 in our direct product group. Since $a \not\in H$, clearly, $a + H$ is a distinct element of $\Bbb Z_6 \times \Bbb Z_4/H$ from $H$. So we have as our second element of the quotient group:

$(0,1) + H = \{(0,1),(2,3),(4,1),(0,3),(2,1),(4,3)\}$.

Unfortunately, we see that $2a = 2(0,1) = (0,2) \in H$, which means that $(0,1) + H$ has order 2 in the quotient group.

However, clearly $(1,0) \not\in H$ nor $(0,1) + H$, so $(1,0) + H$ is a 3rd element of the quotient group:

$(1,0) + H = \{(1,0),(3,2),(5,0),(1,2),(3,0),(5,2)\}$.

Again, $2(1,0) = (2,0) \in H$, so $(1,0) + H$ is of order 2 in the quotient group. This tells us the quotient group is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$, since it has more than one element of order 2.

For completeness' sake, we list the last element of the quotient group (which is the set of all the elements of the original direct product not yet listed in any coset) namely:

$(1,1) + H = \{(1,1),(3,3),(5,1),(1,3),(3,1),(5,3)\}$

It should be clear that the mapping $\phi:\Bbb Z_2 \times \Bbb Z_2 \to \Bbb Z_6 \times \Bbb Z_4/H$ given by:

$\phi(a,b) = (a,b) + H$

is an isomorphism.

Looking at it the other way, you should be able to come up with a surjective homomorphism from $\Bbb Z_6 \times \Bbb Z_4 \to \Bbb Z_2 \times \Bbb Z_2$ with kernel $H$. Can you?
Thanks for the help, Deveno.

However, I am having difficulties with the following issue.

I can see that we need to find an element that is not in H in order to generate the next element of the quotient group, and then look for an element that is in neither of the two discovered groups, and so on.

However, I do not understand your arguments regarding the order of such an element.

For example, you write:

"Since our quotient group has order 4, we might hope that an element $a$ of order 4 in $\Bbb Z_6 \times \Bbb Z_4$ generates the cosets. "

Why, exactly, are you hoping an element a of order 4 in $\Bbb Z_6 \times \Bbb Z_4$.

What exactly is the connection between the order of the element and the process of generating the cosets?

Can you clarify this issue?

Note that I am still reflecting on the issue of the isomorphism and the surjective homomorphism that you mention.

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
There are basically two ways of looking at a quotient group, each of which has its attractions.

1) First way: as a group made out of $H$-sized "chunks" of $G$, where $H$ is a normal subgroup of $G$. We use $H$ to define an equivalence relation on $G$, which partitions $G$ into these "chunks", called "cosets of $H$" (the normality of $H$ ensures that right cosets and left cosets agree). Since we are lumping ELEMENTS of $G$ together into "equivalence classes", it is easy to see we wind up with a group with fewer things in it.

"Naming" the equivalence classes poses a bit of a problem, we have several choices for a name for each coset (picking a representative uniquely determines the COSET, but the representative itself is not unique, any element of the coset could have the coset named after it: there is some unavoidable redundancy).

2) Second way: a quotient group of $G$ is another group $G'$ together with a surjective homomorphism $\phi: G \to G'$. The normal subgroup of $G$ we are "modding out" in this view is the kernel of the homomorphism $\phi$.

The fundamental isomorphism theorem basically says these two ways are equivalent.

Now, given that we have a quotient group of order 4, our fondest wish would be for it to be the nicest possible kind of group of order 4, a cyclic one. Cyclic groups are very nice, and easy to understand, and have an almost transparent inner structure. Since $G$ has elements of order 4, if one of those generated the cosets, it would mean the quotient was cyclic. Unfortunately, that is not the case, here.

Now, given ANY homomorphism $\phi: G \to G'$ between 2 groups, it is ALWAYS the case that:

$|\phi(g)|$ divides $|g|$. So it is a natural question to ask, does:

$|\phi(g)| = |g|$?

When the order of the $g$ involved is small (in this example, we looked at $g$ with an order of 4), often the possibilities for the order of $\phi(g)$ are quite limited, and perhaps can provide useful information.

#### Peter

##### Well-known member
MHB Site Helper
There are basically two ways of looking at a quotient group, each of which has its attractions.

1) First way: as a group made out of $H$-sized "chunks" of $G$, where $H$ is a normal subgroup of $G$. We use $H$ to define an equivalence relation on $G$, which partitions $G$ into these "chunks", called "cosets of $H$" (the normality of $H$ ensures that right cosets and left cosets agree). Since we are lumping ELEMENTS of $G$ together into "equivalence classes", it is easy to see we wind up with a group with fewer things in it.

"Naming" the equivalence classes poses a bit of a problem, we have several choices for a name for each coset (picking a representative uniquely determines the COSET, but the representative itself is not unique, any element of the coset could have the coset named after it: there is some unavoidable redundancy).

2) Second way: a quotient group of $G$ is another group $G'$ together with a surjective homomorphism $\phi: G \to G'$. The normal subgroup of $G$ we are "modding out" in this view is the kernel of the homomorphism $\phi$.

The fundamental isomorphism theorem basically says these two ways are equivalent.

Now, given that we have a quotient group of order 4, our fondest wish would be for it to be the nicest possible kind of group of order 4, a cyclic one. Cyclic groups are very nice, and easy to understand, and have an almost transparent inner structure. Since $G$ has elements of order 4, if one of those generated the cosets, it would mean the quotient was cyclic. Unfortunately, that is not the case, here.

Now, given ANY homomorphism $\phi: G \to G'$ between 2 groups, it is ALWAYS the case that:

$|\phi(g)|$ divides $|g|$. So it is a natural question to ask, does:

$|\phi(g)| = |g|$?

When the order of the $g$ involved is small (in this example, we looked at $g$ with an order of 4), often the possibilities for the order of $\phi(g)$ are quite limited, and perhaps can provide useful information.
Thanks for the help, Deveno.

However, I do not think I have fully understood all the implications regarding the order of elements in this context.

Indeed, I was just reading Beachy and Blair: Abstract Algebra, Example 3.8.12 (page 174) which reasons from the order of elements of a factor group to an isomorphism - can you please help me see their argument.

The example (which is extremely similar to the example we have been talking about) reads as follows: (See Attachment)

==============================================================

Example 3.8.12

Let $$\displaystyle G = \mathbb{Z}_4 \times \mathbb{Z}_4$$

and let $$\displaystyle N = \{ (0,0), (2,0), (0,2), (2,2) \}$$

(We have omitted the brackets denoting congruence classes because that makes the notation too cumbersome)

There are fur cosets of this subgroup, which we can choose as follows:

$$\displaystyle N, \ \ (1,0) + N , \ \ (0,1) + N , \ \ (1,1) + N$$

The representatives of the cosets have been carefully chosen to show that each non-trivial element of the factor group has order 2, making the factor group $$\displaystyle G/N$$ isomorphic to $$\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$$.

... ... etc etc (see attachment)

===============================================================

I do not follow the logic that allows us to argue that each non-trivial element of the factor group has order 2, making the factor group $$\displaystyle G/N$$ isomorphic to $$\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$$.

Can you explain ... and further, can you formally and rigorously prove this implication.

By the way the example then goes on to another factor group and argues another isomorphism based on the cyclic nature of the factor group - see attachment - but I am puzzled by this reasoning as well.

Hoping that someone can help.

Peter

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
Coset multiplication (here I use multiplication in the general sense of "group operation") is easy:

$(xH)\ast(yH) = (x\ast y)H$.

In other words, the product of the cosets $xH$ and $yH$ is the coset containing $x\ast y$.

Now in the example you have given, we have:

$[(1,0) + N] + [(1,0) + N] = [(1,0) + (1,0)] + N = (2,0) + N = N$

(since (2,0) is an element of $N$).

Thus in the quotient group $G/N$ we have an element $x \neq e$ with:

$x\ast x = x^2 = e$, showing $|x| = 2$ (remember $N$ is the identity of $G/N$).

Now for ANY element $g \in G$ with $|g| = k$, it is certainly true that:

$g^k = e$, so all the more so it will be true in $G/N$ that:

$(gN)^k = (g^k)N = eN = N$. But it may happen that for some smaller integer $t$ we have:

$(gN)^t = N$.

For any group, however, it is easy to show that if $x \in G$, and:

$x^m = e$, with $|x| = s$ we must have that $s$ is a divisor of $m$.

Why? because if we write:

$m = qs + r$, with either $r = 0$ or $r < s$, we get:

$e = x^m = x^{qs + r} = (x^{qs})(x^r) = (x^s)^q(x^r) = (e^q)(x^r) = ex^r = x^r$.

If $0 < r < s$, this contradicts the fact that $|x|$ is the least positive integer with:

$x^s = e$, forcing us to conclude that $m = qs$ that is: $s|m$.

#### Peter

##### Well-known member
MHB Site Helper
Coset multiplication (here I use multiplication in the general sense of "group operation") is easy:

$(xH)\ast(yH) = (x\ast y)H$.

In other words, the product of the cosets $xH$ and $yH$ is the coset containing $x\ast y$.

Now in the example you have given, we have:

$[(1,0) + N] + [(1,0) + N] = [(1,0) + (1,0)] + N = (2,0) + N = N$

(since (2,0) is an element of $N$).

Thus in the quotient group $G/N$ we have an element $x \neq e$ with:

$x\ast x = x^2 = e$, showing $|x| = 2$ (remember $N$ is the identity of $G/N$).

Now for ANY element $g \in G$ with $|g| = k$, it is certainly true that:

$g^k = e$, so all the more so it will be true in $G/N$ that:

$(gN)^k = (g^k)N = eN = N$. But it may happen that for some smaller integer $t$ we have:

$(gN)^t = N$.

For any group, however, it is easy to show that if $x \in G$, and:

$x^m = e$, with $|x| = s$ we must have that $s$ is a divisor of $m$.

Why? because if we write:

$m = qs + r$, with either $r = 0$ or $r < s$, we get:

$e = x^m = x^{qs + r} = (x^{qs})(x^r) = (x^s)^q(x^r) = (e^q)(x^r) = ex^r = x^r$.

If $0 < r < s$, this contradicts the fact that $|x|$ is the least positive integer with:

$x^s = e$, forcing us to conclude that $m = qs$ that is: $s|m$.
OK, thanks, follow what you have said ... (forgive me if I am being slow ...) but still need help with the link between the order of the elements in the factor group and the isomorphism.

Mind you, one can "see" the isomorphism $$\displaystyle G/N \cong \mathbb{Z}_2 \times \mathbb{Z}_2$$, for example

$$\displaystyle (1,0)H \to (1,0)$$ etc

BUT

... what exactly is the logic connecting the order of elements in the factor group and the isomorphism.

If we have two groups of the same order, with each of the elements having the same order, can we claim they are isomorphic, for example?

Can you help?

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
Well, in general, no.

Two isomorphic groups will, of course, have elements of the same order(s), but it IS possible to have two non-isomorphic groups of the same order, with the same number of elements of each possible order.

The smallest example I can think of is these 2 groups of order 27:

$G = \Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_3$

$G' = \left\{\begin{bmatrix}1&a&b\\0&1&c\\0&0&1 \end{bmatrix}: a,b,c \in \Bbb Z_3\right\}$

each of which has 26 elements of order 3, but the first is abelian, while the second is not.

#### Peter

##### Well-known member
MHB Site Helper
Well, in general, no.

Two isomorphic groups will, of course, have elements of the same order(s), but it IS possible to have two non-isomorphic groups of the same order, with the same number of elements of each possible order.

The smallest example I can think of is these 2 groups of order 27:

$G = \Bbb Z_3 \times \Bbb Z_3 \times \Bbb Z_3$

$G' = \left\{\begin{bmatrix}1&a&b\\0&1&c\\0&0&1 \end{bmatrix}: a,b,c \in \Bbb Z_3\right\}$

each of which has 26 elements of order 3, but the first is abelian, while the second is not.

Ok, thanks ... but the link between the order of the elements in the factor group and Beachy and Blair's argument in Example 3.8.12, page 174 (see attachment) remains a mystery to me ...

To restate my problem, I do not follow the logic that allows Beachy and Blair to argue that because each non-trivial element of the factor group has order 2, the factor group $$\displaystyle G/N$$ is isomorphic to $$\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$$.

Can you please clarify the basis of their argument ... they seem to argue the existence of an isomorphism from the fact that each non-trivial member of the factor group has order2 ... I'm perplexed ...

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
This has to due with the rather limited possibilities for a group of order 4:

1) Any group of order 4 is abelian (can you prove this?).

2) If a group of order 4 has no element of order 4 (if it does, it is cyclic of order 4) then every non-identity element is of order 2. We can thus write it as:

$G = \{e,a,b,ab\}, a^2 = b^2 = (ab)^2 = e$.

The last equation:

$(ab)^2 = e$ together with $a^2 = b^2 = e$ gives:

$e = (ab)^2 = abab$
$a = a(abab) = a^2(bab) = e(bab) = bab$
$ab = (bab)b = (ba)b^2 = (ba)e = ba$.

If we define:

$\phi:G \to \Bbb Z_2 \times \Bbb Z_2$ by:

$\phi(e) = (0,0)$
$\phi(a) = (1,0)$
$\phi(b) = (0,1)$
$\phi(ab) = (1,1)$

It is clear to see that:

$\phi(ab) = (1,1) = (1,0) + (0,1) = \phi(a) + \phi(b)$

(the other 15 products are trivial to verify: we can use the fact that both groups are abelian to reduce the products we have to check to 10, we just checked 1, which leaves 9, 4 of these involve the identity, and are almost immediate, of the remaining 5, 3 can be deduced from order arguments, leaving just:

$\phi((ab)a) = \phi(ab) + \phi(a)$
$\phi((ab)b) = \phi(ab) + \phi(b)$

that actually require any thought).

Having displayed a bijective homomorphism, we conclude $(G,\ast) \cong (\Bbb Z_2 \times \Bbb Z_2,+)$

It often turns out that investigating the elements of order 2 is quite illuminating in determining which group we have of a given order. This is always the case if $|G|$ is a power of 2. In general, if $p$ is prime and $p$ divides the order of a finite group $G$, looking at the number of elements of order $p$ can help determine which isomorphism class $G$ belongs to (but as the example I gave above shows, does NOT completely determine $G$).

Completely determining how many group isomorphism classes we have of a group of order $n$ there are is, in general, a difficult problem (but, not insurmountable...this problem HAS been solved, but the amount of calculation required is truly mind-boggling). Fortunately, you are rarely presented with a "bad" example (such as determining which group of order 1024 you have...there are almost 50 billion of them!!!...fortunately only around 400 million (!) of those are "common").

But for groups of order less than 32, this problem is simple enough that you could solve it yourself (you might struggle with the groups of order 16...there are more than you might think).

#### Peter

##### Well-known member
MHB Site Helper
This has to due with the rather limited possibilities for a group of order 4:

1) Any group of order 4 is abelian (can you prove this?).

2) If a group of order 4 has no element of order 4 (if it does, it is cyclic of order 4) then every non-identity element is of order 2. We can thus write it as:

$G = \{e,a,b,ab\}, a^2 = b^2 = (ab)^2 = e$.

The last equation:

$(ab)^2 = e$ together with $a^2 = b^2 = e$ gives:

$e = (ab)^2 = abab$
$a = a(abab) = a^2(bab) = e(bab) = bab$
$ab = (bab)b = (ba)b^2 = (ba)e = ba$.

If we define:

$\phi:G \to \Bbb Z_2 \times \Bbb Z_2$ by:

$\phi(e) = (0,0)$
$\phi(a) = (1,0)$
$\phi(b) = (0,1)$
$\phi(ab) = (1,1)$

It is clear to see that:

$\phi(ab) = (1,1) = (1,0) + (0,1) = \phi(a) + \phi(b)$

(the other 15 products are trivial to verify: we can use the fact that both groups are abelian to reduce the products we have to check to 10, we just checked 1, which leaves 9, 4 of these involve the identity, and are almost immediate, of the remaining 5, 3 can be deduced from order arguments, leaving just:

$\phi((ab)a) = \phi(ab) + \phi(a)$
$\phi((ab)b) = \phi(ab) + \phi(b)$

that actually require any thought).

Having displayed a bijective homomorphism, we conclude $(G,\ast) \cong (\Bbb Z_2 \times \Bbb Z_2,+)$

It often turns out that investigating the elements of order 2 is quite illuminating in determining which group we have of a given order. This is always the case if $|G|$ is a power of 2. In general, if $p$ is prime and $p$ divides the order of a finite group $G$, looking at the number of elements of order $p$ can help determine which isomorphism class $G$ belongs to (but as the example I gave above shows, does NOT completely determine $G$).

Completely determining how many group isomorphism classes we have of a group of order $n$ there are is, in general, a difficult problem (but, not insurmountable...this problem HAS been solved, but the amount of calculation required is truly mind-boggling). Fortunately, you are rarely presented with a "bad" example (such as determining which group of order 1024 you have...there are almost 50 billion of them!!!...fortunately only around 400 million (!) of those are "common").

But for groups of order less than 32, this problem is simple enough that you could solve it yourself (you might struggle with the groups of order 16...there are more than you might think).
Thank you for that most informative post ... really helpful!

Will now work through it carefully.

Peter

#### Peter

##### Well-known member
MHB Site Helper
Thank you for that most informative post ... really helpful!

Will now work through it carefully.

Peter
Hi Deveno,

Can you provide some guidance on the approach to proving that any group of order 4 is abelian?

Would appreciate some help,

Peter

#### Deveno

##### Well-known member
MHB Math Scholar
Let $G$ be a finite group with 4 elements. One of these is, of course, the identity. We can write $G$ as:

$G = \{e,x,y,z\}$.

Some trial-and-error quickly establishes that the element $xy$ (which must lie in $G$, by closure) must either be $e$ or $z$.

Case 1) $xy = e$.

In this case, neither $x$ nor $y$ can be of order 2, since any element of order 2 is its own inverse. Since (by Lagrange) the order of each of these must divide 4, and 1 and 2 are not possible, they must both be of order 4, hence $G$ is cyclic, and thus abelian.

Case 2) $xy = z$. We can thus write:

$G = \{e,x,y,xy\}$.

Now consider: $yx$ must also lie in $G$.

$yx = e \implies x = y^{-1} \implies xy = e$, contradiction.

$yx = x \implies y = e$, contradiction.

$yx = y \implies x = e$, contradiction.

Hence the only possibility is $yx = xy$, thus:

$e$ commutes with everything,

$x$ commutes with everything,

$y$ commutes with everything, so all elements commute with all others. Thus $G$ is abelian.