If $|G|>1$ is not prime, there exists a subgroup of $G$ which is not $G$ or $e$.

Question

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.

Answer 1

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 $b$.

Answer 2

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 $G$.

Edit:

$|G|$ 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.

Answer 3

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

Answer 4

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.

Answer 5

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.

Answer 6

I would start with stating that for $g \in G$ we know that $\langle g \rangle \leq G$. At this stage, we either have $\langle g \rangle < G$ is a proper subgroup as you desire, or $G$ is cyclic and $\langle g \rangle = G$.

Let us denote $|G|=mn$ for $m,n >1$. In the case where $G$ is cyclic, then consider the element $g^m \in G$. We can now generate a new cyclic subgroup with this element, $\langle g^m \rangle$.

It is known that $g^{|G|} = e \;\;\forall g \in G$. Thus,

$$g^{mn}=e \implies (g^m)^n=e \implies o(g^m)=n$$

We also know that the order of a cyclic subgroup is equal to the order of the generating element, thus:

$$|\langle g^m \rangle|=n<|G|$$

Now, we have a proper cyclic subgroup strictly not equal to $G$, thus you have proved what was asked of you.