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

Question

Question: Prove the following: if|G|>1 is not prime, then $G$ has a subgroup other than $G$ and {e} We know that $\langle g \rangle$ is a subgroup of $G$ by previous part of the question and $G$ is finite

From other similar question on here I found this:

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.

Does it mean that $\langle g^m \rangle$ and $\langle g^n \rangle$ are also subgroups and is that the proof?

Answer 1

I'm not very sure of what you are asking exactly. I'll prove the theorem, and I think that your question is answered in the proof. I hope you find it useful.

Take any $g\in G$, $g\neq e$. If $g$ does not generate $G$, $\langle g\rangle$ is a subgroup and it is not $G$ or $\{e\}$. If $g$ generates $G$ then $G$ is cyclic and the order of $g$ is $|G|$. Write $|G|=mn$ with $m,n>1$. Then $\langle g^m\rangle$ and $\langle g^n\rangle$ are proper subgroups. This completes the proof.

Answer 2

Exactly.

The final argument may run as follow: take $g\neq e$ so that $|\langle g\rangle|\neq 1$. If $\langle g\rangle\neq G$ you have nothing to prove.

If $\langle g\rangle=G$ then consider $g^d$ for some proper divisor of $|G|$. Then by your argument the subgroup generated by $g^d$ is non-trivial and proper.

Answer 3

Suppose $\lvert G\rvert=rs\enspace(r,s>1)$ and take $g\in G,\enspace g\ne e$.

• If $\langle\, g\,\rangle\neq G$, very well: you have your proper subgroup.
• If $\langle\, g\,\rangle= G $, $G$ is cyclic, and $g^r$ has order $s$, hence generates a proper subgroup.