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My question is in the title: Prove this statement.

Immediately before this question, I proved that $\langle g \rangle$ is a subgroup of $G$. The question specifies $G$ is a finite group.

My attempt:

By the previous question, provided that g is in G, then $\langle g \rangle$ is a subgroup of G. $\langle g\rangle$ is not the identity nor is it G itself unless G = $\langle g \rangle$ (in which case G is cyclic). So we can write $G = ${$e,g,g^2,...,g^{o(g)-1}$}. If G is cyclic, it's order may be prime or non prime. For the purposes of the question, say |G| is nonprime. Then we can write a subgroup as {$g^n$}, where $0<n<o(g)$ so as to exclude the trivial case. Done (I think).

If |G| ≠ |$\langle g \rangle$|, and in particular, |G| < |$\langle g \rangle$| by Lagrange, then |G| is not prime (since it is not equal to it's cyclic group) and so $\langle g \rangle$ qualifies as a subgroup.$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\blacksquare$

Is it okay?

MASSIVE EDIT: Question should say |G|$> 1$ is NOT prime. Apologies.

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The statement in the title is wrong. Is it a typo? (Side note: Many prefer that the content of the question doesn't depend on the title. The title draws people in, hopefully gives a good idea of the content, but the body contains the full context, ideally.) – Jonas Meyer Jan 30 '13 at 16:47
Thanks Jonas. I saw it just before your comment and put an edit into the content. – CAF Jan 30 '13 at 16:50
Please do edit ASAP your post's title. – DonAntonio Jan 30 '13 at 16:55
up vote 4 down vote accepted

If $|G|>1$ is not prime, say $|G|=ab$ with $a,b>1$. Select $g\in G\setminus\{1\}$. If $\langle g\rangle \ne G$, we are done. Otherwise, the order of $g$ is $ab$ and we can consider $\langle g^a\rangle$ instead, which has order $b$.

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Could you elaborate on how if the order of G is $ab$ then a subgroup is necessarily $\langle g^a \rangle$ which has order $b$? Is it a consequence of Lagrange, i.e the order has to divide the order of G? Why must we have $\langle g^a \rangle$ then? – CAF Jan 30 '13 at 17:01
Actually I think I understand. If $<g^a>$ has order $b$ then $<(g^a)^b> = <(g^{ab})> = <(g^{|G|})> = e$. (One element must have an order |G|). Is this correct? – CAF Jan 30 '13 at 17:25

Hint:Let $o(G)=mn$ then one subgroup is $<g^m>$.

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Looking at the subgroup $H$ generated by an element $g\ne e$ is a good idea. Perhaps this group is a proper subgroup. Then we are finished.

If $H$ is all of $G$, let $|G|=ab$ where neither $a$ nor $b$ is $1$, and consider the group generated by $g^a$. That seems to be what you were thinking of, but one needs to prove that under these conditions the group generated by $g^a$ is indeed a proper subgroup.

If instead we consider the subgroup generated by $g^n$, where we only specify that $1\lt n\lt |G|$, then this subgroup is often all of $G$. In fact it is all of $G$ precisely if $|G|$ and $n$ are relatively prime.

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Okay thanks. One question just to check I am not confused: G can only have one cyclic group right? The $<g^a>$ above is a cyclic group but because it's order is less than the order of <g> it is not equal to <g> – CAF Jan 30 '13 at 19:20
I don't know what ou mean. A group can have many cyclic subgroups. This discussion is getting kind of long, the system objects. I will delete most of my previous comments, suggest you do the same. That will leave room for more, if necessary! – André Nicolas Jan 30 '13 at 19:22
A group can have many cyclic subgroups, I understand this, but say, when I write G = <g>, is this <g> unique? – CAF Jan 30 '13 at 19:34
Also, when in the proof did I make explicit that |G| was non-prime? – CAF Jan 30 '13 at 19:37
Well, you didn't, but the your original proof is not correct. Remember we only need to worry if $G$ is cyclic with generator $g$. Then we need to produce a proper subgroup. So we are looking at the case $|G|=m$, where $m$ is the order of $g$. We use the fact $m$ is not prime by writing $m=ab$ where neither $a$ nor $b$ is $1$. Then we prove that the group generated by $g^a$ is a proper subgroup of the group generated by $g$. – André Nicolas Jan 30 '13 at 19:59

You write:

  • $\langle g\rangle$ is not the identity nor is it $G$ itself unless $G = \langle g\rangle$

but then you never address the case that $G = \langle g\rangle$. Why do you assume that this is not the case?

In fact the statement in your title is false, if $|G| > 1$ is prime then $G \simeq \mathbb Z/p$ and any non-identity element generates all of $G$.


$|G|$ not prime makes much more sense as a question :)

With your edit, the problem I still see in your proof is that you say $\langle g^n\rangle \neq G$ is a subgroup for $n < |G|$ but this is not true for any arbitrary $n$. You have to show how to pick such an $n$ and why that particular $n$ works.

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@CAF: Do your claim in the title, for $G=\{0,1\}$. Can you get any other proper subgroup? ;-) – Babak S. Jan 30 '13 at 16:51
Hi Jim. I will correct the title but I put an edit in the content. It should say |G| > 1 is NOT prime. In terms of addressing $G = \langle g \rangle$, I said if this was the case then any element in $<g>$(=G) can form a subgroup. Is there something else I should add?. – CAF Jan 30 '13 at 16:54
Yes, when $G$ is cyclic you should say how to pick $g$, because for some $g$ you will get $\langle g \rangle = 1$ or $= G$. – Jim Jan 30 '13 at 17:05
Did I not write that $0<n<o(g)$, i.e exclude the identity and G itself? – CAF Jan 30 '13 at 17:11
You did write that but it's not correct. For example assume $G = \langle g\rangle$ is cyclic and $|G| = 4$. Then $\langle g^3\rangle = G$ even though $0 < 3 < 4$. – Jim Jan 30 '13 at 17:15

Just for clarity's sake, here is the proof that if $g$ generates $G$, and $|G| = mn,\ 1 < m,n < |G|$, that $1 < |\langle g^m \rangle| < |G|$.

If $|\langle g^m \rangle| = 1$ then $g^m = e$ contradicting that $|g| = |G| = mn$.

Since $e = g^{mn} = (g^m)^n$, we must have $|g^m| \leq n < mn = |G|$.

One can indeed prove that $|g^m| = n$ but this is not necessary.

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