Determine the subgroup

[GreenGoblin]

Determine the subgroups

Let S: = {(123),(235)} be a subset of $\sum_{5}$. Determine the subgroup <S> of $\sum_{5}$?

What exactly is this asking? Is S the set of just two elements here? Or all elements containing these permutations? If the former how can S be a subgroup?

Let G = <g> be a cyclic group of order 12. Find all subgroups of G.

Again, I am not 100% clear on the notation and just want an idea of what to do. The terms seem so broad.

Gracias,
GreenGoblin

 

[Evgeny.Makarov]

Re: Determine the subgroups

I recommend going back to your sources (textbook or lecture notes) for notations and definitions. I think that the concepts from the first question are described in this Wiki page. In the second question, $G = \{e, g,g^2,\dots,g^{11}\}$.

 

[GreenGoblin]

Re: Determine the subgroups

I am not learning from 'textbook or lecture notes'. Please do not make these assumptions.

Anyway, I do know the definitions, it's just it's my first attempt on these kind of problems and I do not fully know how to solve the problems. Here is a more direct question I do not have an answer for: I know the product of two elements in a group must be in the group. But is it a requisite for the product of an element times itself to be in the group? i.e. gg must be in G?

 

[hmmm16]

Re: Determine the subgroups

Yes, from what you just said. Let $$ G $$ be a group and $$ g\in G $$ and $$ g\in G $$ so $$ g.g\in G $$

 

[GreenGoblin]

Re: Determine the subgroups

So if for example $g^{2}$ is in the group $g^{4}$ must be and thus forth? $<e,g,g^{11}>$ for example can't be a subgroup since it doesn't meet that condition. It meets all others.

But that means the only subgroup is G itself.

 

[Evgeny.Makarov]

Re: Determine the subgroups

I am not learning from 'textbook or lecture notes'. Please do not make these assumptions.

Well, you must be learning from some sources. Since you said you were not clear on the notation, I thought it would be a good idea to go back where the notation was introduced, especially since notations and definitions often vary between textbooks and courses. In my experience, failing to sufficiently absorb definitions is a major source of questions and misunderstandings for beginner mathematicians.

I know the product of two elements in a group must be in the group. But is it a requisite for the product of an element times itself to be in the group?

The fact that the product of two elements in a group must be in the group is just an expression in plain English (which makes it less precise) of one of the components of the definition of a group. Namely, $(G,{\cdot})$ is a group if G is a set and ${\cdot}:G\times G\to G$ is a binary operation (i.e., function) on G. The definition of a function, in turn, implies that $\cdot$ converts any element of the domain $G\times G$, i.e., any pair (g, h), whether g and h are the same elements of G or not, into an element of G.

So if for example $g^{2}$ is in the group $g^{4}$ must be and thus forth?

Yes.

$<e,g,g^{11}>$ for example can't be a subgroup since it doesn't meet that condition. It meets all others.

I agree that the set $\{e,g,g^{11}\}$ is not a subgroup because it is not closed under the operation. However, $\langle e,g,g^{11}\rangle$ by definition is the subgroup generated by these elements. Since the elements include g, this subgroup coincides with G.

But that means the only subgroup is G itself.

What about $\langle g^3\rangle=\{e,g^3,g^6,g^9\}$?

 

[Deveno]

Re: Determine the subgroups

Let S: = {(123),(235)} be a subset of $\sum_{5}$. Determine the subgroup <S> of $\sum_{5}$?

What exactly is this asking? Is S the set of just two elements here? Or all elements containing these permutations? If the former how can S be a subgroup?

Let G = <g> be a cyclic group of order 12. Find all subgroups of G.

Again, I am not 100% clear on the notation and just want an idea of what to do. The terms seem so broad.

Gracias,
GreenGoblin

for your first question, i believe the group in question is $S_5$, the symmetric group on 5 "letters". note that for any subset S of a group G, if a is in S, then <a> is a subgroup of <S>. so the subgroup generated by (1 2 3) and (2 3 5) in $S_5$, contains the elements {e, (1 2 3), (1 3 2), (2 3 5), (2 5 3)} at a minimum.

another fact is that if a,b are BOTH in S, then ab is in <S>, as well as ba, and products such as:

$aba, a^2b^{-1}a, (ab)^2$ etc.

so even a subgroup generated by just 2 elements, can be rather large.

to find the subgroup you are looking for, note that neither element of S moves 4, so no element of <S> will move 4. thus |<S>| $\leq |S_4|$ = 24 (<S> only permutes elements of {1,2,3,5}). in fact, <S> must only contain even permutations (permutations that can be written as an even number of transpositions), since both generators are 3-cycles. the order of <S> is, in fact, a divisor of 24 that is divisible by 3. try to prove that |<S>| > 6, for starters (we've already listed 5 elements, show that (1 2 3)(2 3 5) and (2 3 5)(1 2 3) give at least 2 more). do you know what an alternating group is?

********

for your second problem, it might help to look at a "concrete" cyclic group of order 12, $\mathbb{Z}_{12}$, the integers modulo 12.

 

[GreenGoblin]

Re: Determine the subgroups

for your first question, i believe the group in question is $S_5$, the symmetric group on 5 "letters". note that for any subset S of a group G, if a is in S, then <a> is a subgroup of <S>. so the subgroup generated by (1 2 3) and (2 3 5) in $S_5$, contains the elements {e, (1 2 3), (1 3 2), (2 3 5), (2 5 3)} at a minimum.

another fact is that if a,b are BOTH in S, then ab is in <S>, as well as ba, and products such as:

$aba, a^2b^{-1}a, (ab)^2$ etc.

so even a subgroup generated by just 2 elements, can be rather large.

to find the subgroup you are looking for, note that neither element of S moves 4, so no element of <S> will move 4. thus |<S>| $\leq |S_4|$ = 24 (<S> only permutes elements of {1,2,3,5}). in fact, <S> must only contain even permutations (permutations that can be written as an even number of transpositions), since both generators are 3-cycles. the order of <S> is, in fact, a divisor of 24 that is divisible by 3. try to prove that |<S>| > 6, for starters (we've already listed 5 elements, show that (1 2 3)(2 3 5) and (2 3 5)(1 2 3) give at least 2 more). do you know what an alternating group is?

********

for your second problem, it might help to look at a "concrete" cyclic group of order 12, $\mathbb{Z}_{12}$, the integers modulo 12.

I don't think it is true that only permutations which fix 4 can be included. I think the set is the set of all permutations which either contain (123) or (235). Thus (123) and (123)(45), etc, would be included?

 

[Evgeny.Makarov]

Re: Determine the subgroups

I don't think it is true that only permutations which fix 4 can be included.

So, you think that some permutations that move 4 are in <S>. Could you give an example of such permutation and show how it is expressed through (123) and (235)? If neither of these two permutations (nor their inverses) move 4, how can a product of them move 4?

I think the set is the set of all permutations which either contain (123) or (235).

What does it mean for one permutation to contain another one?

Thus (123) and (123)(45), etc, would be included?

No, because (123)(45) moves 4.

 

[GreenGoblin]

Re: Determine the subgroups

So, you think that some permutations that move 4 are in <S>. Could you give an example of such permutation and show how it is expressed through (123) and (235)? If neither of these two permutations (nor their inverses) move 4, how can a product of them move 4?

What does it mean for one permutation to contain another one?

No, because (123)(45) moves 4.

Contains as in contains that cycle. That's what it's asking. The subgroup found from the set of permutations that contain those cylces. (123)(45) contains (123), because it moves 1 to 2, 2 to 3, and 3 to 1. It wants all the permutations that have this property (plus there inverses and products). Likewise for (235).

The set doesn't mean just the permutations (123) and (235). It means the set of all permutations (in sigma 5) that contain these cycles.

 

[Evgeny.Makarov]

Re: Determine the subgroups

With all due respect, your definition of <S>, which involves the concept "contains," sounds suspicious because it differs from the definitions in Wikipedia (link in post #2) and PlanetMath. Also, I've never encountered "contains" in the context of permutations. Could you check the definition of <S> in your sources?

So, you seem to be claiming that f contains g if $f = g\cdot h=h\cdot g$ for some h such that the sets of elements that g and h move are disjoint. Note that defining f to contain g if $f = g\cdot h$ for some h does not make much sense because then every permutation contains every other permutation.

 

[GreenGoblin]

Re: Determine the subgroups

With all due respect, your definition of <S>, which involves the concept "contains," sounds suspicious because it differs from the definitions in Wikipedia (link in post #2) and PlanetMath. Also, I've never encountered "contains" in the context of permutations. Could you check the definition of <S> in your sources?

So, you seem to be claiming that f contains g if $f = g\cdot h=h\cdot g$ for some h such that the sets of elements that g and h move are disjoint. Note that defining f to contain g if $f = g\cdot h$ for some h does not make much sense because then every permutation contains every other permutation.

The set of all permutations that have the effect of either (123) or (235). Thus literally just (123), (123)(45), (235) and (235)(14). And then, the products/inverses etc of them, plus the identity, would form the neccessary subgroup.

I don't see what's not to get about this definition? As I said (123) being in the set means all permutations sending 1 to 2, 2 to 3 and 3 to 1 are in the set (not just 'the' permutation (123). (123) isn't referring to a permutation here, itself referring to a property of the permutations in the set). There are only two of these so it only adds one more. Since we are s5 this only adds one permutation to the set for each of (123) and (235), were it s6 it would add 5 more for each.

 

[Deveno]

Re: Determine the subgroups

The set of all permutations that have the effect of either (123) or (235). Thus literally just (123), (123)(45), (235) and (235)(14). And then, the products/inverses etc of them, plus the identity, would form the neccessary subgroup.

I don't see what's not to get about this definition? As I said (123) being in the set means all permutations sending 1 to 2, 2 to 3 and 3 to 1 are in the set (not just 'the' permutation (123). (123) isn't referring to a permutation here, itself referring to a property of the permutations in the set). There are only two of these so it only adds one more. Since we are s5 this only adds one permutation to the set for each of (123) and (235), were it s6 it would add 5 more for each.

no, (1 2 3)(4 5) is NOT in the subgroup <S>. it is not a true statment that (1 2 3) is "contained in" (1 2 3)(4 5). (1 2 3) does not represent a "property", it is a certain function $f:X \to X$ where $X = \{1,2,3,4,5\}$, namely:

f(1) = 2
f(2) = 3
f(3) = 1
f(4) = 4
f(5) = 5.

now, while it is true we can regard a function as a subset of X x X, it still is NOT the case that (1 2 3) is a subset of (1 2 3)(4 5). the former contains the pair (4,4), and the latter does not.

i think you have some basic misconceptions about what the nature of a group, and a permutation group are, and you seem reluctant to let other people assist you.

 

[GreenGoblin]

Re: Determine the subgroups

no, (1 2 3)(4 5) is NOT in the subgroup <S>. it is not a true statment that (1 2 3) is "contained in" (1 2 3)(4 5). (1 2 3) does not represent a "property", it is a certain function $f:X \to X$ where $X = \{1,2,3,4,5\}$, namely:

f(1) = 2
f(2) = 3
f(3) = 1
f(4) = 4
f(5) = 5.

now, while it is true we can regard a function as a subset of X x X, it still is NOT the case that (1 2 3) is a subset of (1 2 3)(4 5). the former contains the pair (4,4), and the latter does not.

i think you have some basic misconceptions about what the nature of a group, and a permutation group are, and you seem reluctant to let other people assist you.

No, what I have written is what is being asked for, in this case. It may be an unfamiliar/unconventional notation, but it is what is being asked for. As I said, in the question (123) and (235) are not expressing permutations but rather cycles. It is asking for all permutations that contain these cycles, and thus the group generated by these (four not two) permutations.

 

[Deveno]

Re: Determine the subgroups

the trouble i have with what you are saying is this:

(1 2 3) IS a permutation: cycles are permutations. cycle notation is a shorthand for the permutation:

$\begin{pmatrix}1&2&3&4&5\\2&3&1&4&5 \end{pmatrix}$

now some texts use (a,b,c...) to denote the image set of the ordered tuple (1,2,3...), but in such a case (with $S_5$) one has to list all 5 images to unambigously define what you mean.

i have not seen any indication in this thread that you understand the concept of "the subgroup generated by the set S". and if what you are learning from contains a substantially different definition than that wriiten in say: wikipedia, Fraleigh, Dummit&Foote, Herstein, etc. you are going to have an uphill battle communicating "what you want" with other people. either you're leaving something out in what you're telling us, or you are just mis-understanding the basic concepts involved. i believe the latter is more likely.

(1 2 3)(4 5) does not have 'the same effect" as (1 2 3). (1 2 3)(4 5) affects all 5 elements of its domain, it has no fixed points. (1 2 3) has 2 fixed points, 4 and 5.
(1 2 3)(1 2 3) also has 2 fixed points, 4 and 5 again.

the set $H =\{\sigma \in S_5: \sigma(4) = 4\}$ is a subgroup of $S_5$: in particular, it is closed under composition. ANY two elements of H, when composed, will yield another element of H. the set S = {(1 2 3), (2 3 5)} is a subset of H, and thus <S> (the subgroup generated by S) is a subgroup of H. i'm not making this stuff up, it's fairly standard notation/practice. i have seen little evidence that you know what you're talking about, and several other posters in this thread seem to feel the same.

 

[GreenGoblin]

Re: Determine the subgroups

the trouble i have with what you are saying is this:

(1 2 3) IS a permutation: cycles are permutations. cycle notation is a shorthand for the permutation:

$\begin{pmatrix}1&2&3&4&5\\2&3&1&4&5 \end{pmatrix}$

now some texts use (a,b,c...) to denote the image set of the ordered tuple (1,2,3...), but in such a case (with $S_5$) one has to list all 5 images to unambigously define what you mean.

i have not seen any indication in this thread that you understand the concept of "the subgroup generated by the set S". and if what you are learning from contains a substantially different definition than that wriiten in say: wikipedia, Fraleigh, Dummit&Foote, Herstein, etc. you are going to have an uphill battle communicating "what you want" with other people. either you're leaving something out in what you're telling us, or you are just mis-understanding the basic concepts involved. i believe the latter is more likely.

(1 2 3)(4 5) does not have 'the same effect" as (1 2 3). (1 2 3)(4 5) affects all 5 elements of its domain, it has no fixed points. (1 2 3) has 2 fixed points, 4 and 5.
(1 2 3)(1 2 3) also has 2 fixed points, 4 and 5 again.

the set $H =\{\sigma \in S_5: \sigma(4) = 4\}$ is a subgroup of $S_5$: in particular, it is closed under composition. ANY two elements of H, when composed, will yield another element of H. the set S = {(1 2 3), (2 3 5)} is a subset of H, and thus <S> (the subgroup generated by S) is a subgroup of H. i'm not making this stuff up, it's fairly standard notation/practice. i have seen little evidence that you know what you're talking about, and several other posters in this thread seem to feel the same.

Under your notation then (which I will take your word as being a 'standard' notation), this should say something along the lines of f(1)=2,f(2)=3 and f(3)=1 then. HOWEVER, please just take it as given that what I am saying is the intended meaning of the question IS the intended meaning. We are looking for the subgroup containing these elements. The use of (123) here does NOT denote the permutation (123) rather all permutations that move 1 to 2, 2 to 3, and 3 to 1. Likewise for (235). I accept that this is not conventional notation (as a few have pointed out), however this IS WHAT IS BEING ASKED FOR! I am certain of this as I have stated now multiple times and this needs not be questioned. With that in mind I would appreciate some help finding this subgroup.

If you are not comfortable with the way the question is asked, please reword it to fit what I am saying and help me with this problem! If you can suggest how what I am asking should be better worded, please put it to me and I will confirm whether this is what's meant. But the point is that I am certain of what is INTENDED by the question whether or not it is worded 'correctly' per convention. The question as written initially may carry a different meaning to you as what I need but hopefully I am clear enough here expressing what I am actually asking for. We are looking for 'the subgroup of sigma 5 generated by the set of elements in sigma 5 with either f1=2,f2=3 and f3=1 OR f2=3,f3=5 and f5=2". Better? So we need these (four not two!) elements, inverses, products etc.

 

[Deveno]

Re: Determine the subgroups

if a notation that is being used is non-standard, we have to know. (1 2 3) has an established meaning, a "convention" if you will. using (1 2 3) to mean:

all x in $S_5$ such that x(1) = 2, x(2) = 3 and x(3) = 1 is possible, but that's not what is commonly meant by <(1 2 3)> = {e,(1 2 3), (1 3 2)}.

moreover, i would further argue that such an interpretation of <(1 2 3)> is not even useful in the context of group theory, because the elements of such a set are not closed under multiplication. to see what i mean:

note that (1 2 3)(4 5)(1 2 3)(1 2 3) = (4 5), so (4 5) is in "the subgroup generated by (1 2 3)" (YOUR definition). but (4 5) doesn't have the desired property of sending 1-->2, 2-->3, 3-->1. if one includes (2 3 5) as another "generator" and thus (2 3 5)(1 4), then we get as additional members:

(2 3 5)(4 5)(2 3 5)(2 3 5) = (2 4)
(4 5)(2 4)(4 5) = (2 5)
(2 3 5)(1 4)(2 3 5)(2 3 5) = (1 4)
(2 3 5)(2 5)(2 3 5)(2 3 5) = (2 3)
(2 3)(2 4)(2 3) = (3 4)
(2 4)(1 4)(2 4) = (1 2)
(2 3)(1 2)(2 3) = (1 3)
(4 5)(1 4)(4 5) = (1 5)
(1 5)(1 3)(1 5) = (3 5).

so, in your definition, <S> = $S_5$, because the transpositions generate $S_5$. the usual definition of <S> is: "the smallest subgroup of G containing S", which your group clearly is not (it's all of $S_5$).

again, i concede it's possible to make such a definition, but i can only wonder WHY someone would do so, as its utility is not clear.