Group Theory - Isomorphism question

[StudentR]

GroupTheory - Isomorphisms

Hey I'm stuck on these 2 questions, was wondering if anyone could assist me:

Let G be a nontrivial group.
1) Show that if any nontrivial subgroup of G coincides with G then G is isomorphic to C_p, where p is prime. (C_p is the cyclic group of order p!)

2) Show that if any nontrivial subgroup of G is isomorphic to G then G is isomorphic to Z or C_p, where p is prime. (Z is the set of integers, C_p is the cyclic group of order p!)

Thanks, help would be much appreciated

 

[StudentR]

Hey I'm stuck on these 2 questions, was wondering if anyone could assist me:

Let G be a nontrivial group.
1) Show that if any nontrivial subgroup of G coincides with G then G is isomorphic to C_p, where p is prime. (C_p is the cyclic group of order p!)

2) Show that if any nontrivial subgroup of G is isomorphic to G then G is isomorphic to Z or C_p, where p is prime. (Z is the set of integers, C_p is the cyclic group of order p!)

What I'm stuck on is how to begin with the question. To show that two groups are isomorphic, by definition, we show that the homomorphic map from G to C_p must be bijective. So to begin, do i assume an arbitrary homomorphic map from G to C_p, and then show that it is bijective? Or do i assume to take an element that generates the a subgroup?
I assumed some arbitrary map f from G to C_p, such that f(ab)=f(a)f(b). then f is homomorphic. I then showed that f is injective (1-1) by proving that the kernel of G is the unique element G. Is this ok so far? I am stuck how to prove f is onto.

Help would much be appreciated.

 

[matt grime]

1) I'd prefer to state it as: show that if G has exactly two subgroups then G is (cyclic) of prime order.

Both of these questions are answered by taking any element in G (not the identity) and playing with it (it must generate a subgroup, and that subgroup must be (isomorphic to) G). It would help if you know the theorem that 'if p is a prime that divides |G| there is an element of order p in G'.

 

[StudentR]

thanks for the reply.
what I'm stuck on is how to begin with the question. to show that two groups are isomorphic, by definition, we show that the homomorphic map from G to C_p must be bijective. so to begin, do i assume an arbitrary homomorphic map from G to C_p, and then show that it is bijective? Or do i assume to take an element that generates the a subgroup?

 

[matt grime]

No, you don't have to construct a homomorphism. All you have to do si show that the group is uniquely determined by this property. After all at the momenet you don't even know what the p is, do you?

If I have a group with 6 elements and it is not abelian then it is the symmetric group on 3 objects. If I have a group with 4 elements that is not cyclic then it is C_2 x C_2. If I have a group and its order is a prime I know exactly what group it is. At no point have I had to construct any homomorphisms and check anything is bijective.

It is important you know things like Lagrange's theorem: the order of an element divides the order of the group. And the partial converse: if p is a prime that divides the group order then there is an element of order p. This second one, for instance immediately implies the result in 1) if G is finite. All you have to do is show in 1) that G must be finite.

Then in 2) you can split it into two cases: G finite and G infinite. 1) applies to the first case, again, and the infinite case is straight forward (no element can have finite order, can it?).If you dont' want to apply these theorems, and would rather prove it directly, just pick an element g in G.

In 1) what subgroup does it generate? It must be G, right. Case 1, g has finite order, case 2 g has infinite order, and see what you can deduce.

 

[HallsofIvy]

Please do NOT post the same question more than once. I am merging the thread in "Mathematics- Algebra" into this one.

 

[StudentR]

Is this proof correct..?

(a) First i prove that G is finite. If G is infinite then a 1-1 correspondence between G and C_p is not possible, as eventually, for 2 distinct elements in G, they will be mapped into the same element in C_p, eliminating the 1-1 correspondence that is necessary and sufficient for the isomorphism to hold. Hence G is finite.

Next, take any element g in G such that g is not the identity, and let H=<g> be a subgroup of G (H is a group generated by g). Since G is a finite group, g must ahve finite order.
Now by definition, H is contained in every non-trivial subgroup of G which contains g.
By assumption we take H to coincide with G so that |H|=|G|=|g| (where |a| denotes the order of a).
If p is a prime such that p divides |G| then there exists a g' in G such that |g'|=p , but then g' is in <g>, so that |g|=p, hence |g|=|G|=p.
Thus G is a group of prime order, and hence G is cyclic. (this last statement about the prime order of p was proved elsewhere and can be restated).

(b) For part b of the question we consider 2 cases: G finite and G infinite.

If G is finite then part (a) applies to this case.

If G is infinite then take H=<g> = the set {g^n where n is an element of Z (the integers)} . H is a subgroup of G. Now no element of H can have finite order since G is inifinite. Hence H is isomorphic to Z(the integers) and it follows that G is isomorphic to Z, as wanted.

 

[matt grime]

(a) First i prove that G is finite. If G is infinite then a 1-1 correspondence between G and C_p is not possible, as eventually, for 2 distinct elements in G, they will be mapped into the same element in C_p, eliminating the 1-1 correspondence that is necessary and sufficient for the isomorphism to hold. Hence G is finite.

There is no need to show that an infinite group is not isomorphic to C_p. Notice that you have not used any of the defining property of G, namely that any non-trivial subgroup of G is equal to G. You must show that no ininite group has that property. You should not be even mentioning C_p at this point.

Next, take any element g in G such that g is not the identity, and let H=<g> be a subgroup of G (H is a group generated by g). Since G is a finite group, g must ahve finite order.
Now by definition, H is contained in every non-trivial subgroup of G which contains g.
By assumption we take H to coincide with G so that |H|=|G|=|g| (where |a| denotes the order of a).
If p is a prime such that p divides |G| then there exists a g' in G such that |g'|=p , but then g' is in <g>, so that |g|=p, hence |g|=|G|=p.
Thus G is a group of prime order, and hence G is cyclic. (this last statement about the prime order of p was proved elsewhere and can be restated).

That is a bit too long for my liking. Any non-identity g generates G. If its order is not a prime then its order is p*q and g^q generates a subgroup of G that is not G, which is a contradiction, hence |g|=p a prime, and G=C_p.

(b) For part b of the question we consider 2 cases: G finite and G infinite.

If G is finite then part (a) applies to this case.

If G is infinite then take H=<g> = the set {g^n where n is an element of Z (the integers)} . H is a subgroup of G. Now no element of H can have finite order since G is inifinite.

No, that is not correct. C_2 x Z is an infinite group and it has elements of finite order. Again you have dismissed a case because of some reason that does not use the hypothesis. If G is infinite and G has an element of finite order, h, say, then h generates a finite subgroup, so what is the hypothesis on G?

Hence H is isomorphic to Z(the integers) and it follows that G is isomorphic to Z, as wanted.