# Proof verification: If $G$ is a finite abelian group and $n\mid|G|$ then $G$ has a subgroup of order $n$.

I have been trying to do this exercise and found two answers:

Let $G$ be an abelian group of order $m$. If $n$ divides $m$, prove that $G$ has a subgroup of order $n$.

Showing that a finite abelian group has a subgroup of order m for each divisor m of n

I think I found a clearer way of doing that, and I would like to share it and have it "peer-reviewed".

Lemma: Let $G$ be a finite abelian group of order $m$. Then $\forall g\in G$, $g^m=1$.

Let $G$ be a finite abelian group of order $m$, such that $n\mid m$. Let $g\in G$ be an element of $G$ with (finite) order $O(g)=r$.

Since $g\in G$, then $\langle g\rangle\le G$. By Lagrange's Theorem, then $r\mid m$; then let $m=kr$ for a certain $k$. Thus:

$$g^m=g^{kr}=(g^r)^k=(g^k)^r=1^r=1$$

Proposition: Let $G$ be a finite abelian group of order $m$. If $n\mid m$, then $G$ has a subgroup of order $n$.

If $n\mid m$, let $m=kn$ for a certain $k$. Then for every $g\in G$ (with $g \ne e$, $e$ the identity element of $G$): $$g^m=g^{kn}=1\implies (g^k)^n=1$$

Thus with $h=g^k$, we have $h^n=1$; so the subgroup generated by $h=g^k$ has order $n$.

All you've shown is that the subgroup generated by $h$ has an order that divides $n$, not that it is $n$. For instance, if you had (by mistake) picked the identity element $e$ as $g$ in your proposition, you've have correctly shown that $e^n = e$, but that doesn't mean that $e$ has order $n$; in fact, it has order $1$.
• What if we excluded $e$? Maybe then I could reformulate it to give an inductive argument? Apr 21 '17 at 21:35
• That doesn't work either. For in $Z/32Z$, for instance, you might have $n= 16$, and pick, accidentally, the element $24$. You'd then know that $16 \cdot 24= 0$, but that doesn't make $24$ have order $16$ -- it in fact has order $4$. You've done something good here, but it'll take more than this to produce a proof. Bandaids like "I didn't mean the identity" won't help. And that's (part of) why the usual proof doesn't look like the one you've come up with. Apr 21 '17 at 21:37
• I meant about inducting the argument on the order of the subgroup generated by $h$, but I can see then that my proof would become about the same as the other ones (maybe rephrased, but still). Thanks for your response! Apr 21 '17 at 21:44