# When an Abelian group is cyclic

Let G be a finite abelian group.It contains a non trivial subgroup which is contained in every non trivial subgroup.Then G must be cyclic. This is a problem of Herstein book(Pg 108,#11 2nd edition).I can't solve it Plz help me.

• Suppose $G$ is not cyclic. Then it must be the direct product of two non-trivial subgroups. – almagest Sep 7 '14 at 11:37
• @almagest Quaternion group with 8 elements is not a direct product of its proper subgroups – Panja Sep 7 '14 at 14:25
• Nor is it abelian. – almagest Sep 7 '14 at 14:25
• @Panja, $\;Q_8\;$ is not abelian... – Timbuc Sep 7 '14 at 14:25
• yaa Q8 is not abelian...but then how it follows? – Panja Sep 7 '14 at 14:32

Suppose that $H$ is that unique minimal subgroup of $G$. Then for any $g\in G$:
1. $H\subset \langle g\rangle$ implying $H$ is cyclic,
2. Since $H$ is minimal $|H|=p$ is a prime,
3. $\langle g\rangle$ has order $p^n$ for some $n$,
4. $H=\langle g^{p^{n-1}}\rangle$
Now suppose that $h$ is an element of $G$ with the maximal order $ord(h)=p^m$, then $H=\langle h^{p^{m-1}}\rangle$. For any $g\in G$ with order $p^n\le p^m$, $g^{p^{n-1}}$ is a generator of $H$. So we can assume that $$g^{p^{n-1}}=h^{p^{m-1}}.$$ It follows that $$(gh^{-p^{m-n}})^{p^{n-1}}=1.$$ If $gh^{-p^{m-n}}\ne 1$, then it generates a cyclic group of order at most $p^{n-1}$. This group again contains $H$, we can then proceed by reduction to show that $g$ is a power of $h$.
• $g^{p^{n-1}}$ is an element of order $p$ in $\langle h\rangle$, hence will have the form $h^{kp^{m-1}}$ for some $k$ not divisible by $p$. But then $h^k$ has the same order with $h$ and can play the role of $h$. It may be better just leave it $h^k$. – Quang Hoang Sep 8 '14 at 6:27