TITLE: Let $G$ be a finite group which has a total of no more than five subgroups. Prove that $G$ is abelian.
QUESTION [7 upvotes]: Let $G$ be a finite group which has a total of no more than five subgroups. Prove that $G$ is abelian.

I can prove that if $\left|G\right|\leq5
 $ then $G$ is abelian. Is it equivalent to this problem? If it isn't, give me some hints to prove.
Thanks in advanced.

REPLY [3 votes]: I think this works also.
Suppose that a group $G$ is non-abelian, and choose elements $a$ and $b$ in $G$ such that $ab\neq ba$.  Then the cyclic subgroups $\langle a\rangle$ and $\langle b\rangle$ are distinct, non-trivial proper subgroups of $G$.  Since no group is the union of two proper non-trivial subgroups, there is an element $c\in G\setminus(\langle a\rangle\cup\langle b\rangle)$, and it generates another cyclic subgroup $\langle c\rangle$ distinct from $\langle a\rangle$ and $\langle b\rangle$.  We now have five subgroups $1$, $G$, $\langle a\rangle$, $\langle b\rangle$ and $\langle c\rangle$, so we need only find a sixth.  If any of $a$, $b$ and $c$ has non-prime order, then the cyclic subgroup it generates has a proper, non-trivial subgroup distinct from the first five.  Therefore, we can assume that $a$, $b$ and $c$ each have prime order.
If $G\neq\langle a\rangle\cup\langle b\rangle\cup\langle c\rangle$, we are done as before, since we can choose an element $d\in G\setminus(\langle a\rangle\cup\langle b\rangle\cup\langle c\rangle)$ which generates a new cyclic subgroup.  Therefore, assume that $G = \langle a\rangle\cup\langle b\rangle\cup\langle c\rangle$.
We can assume that at least one of $a$, $b$ and $c$ has order greater than $2$ since, otherwise, $G$ is abelian.  Suppose the order of $a$ is greater than $2$.  Then  $ab, a^{2}b\not\in(\langle a\rangle\cup\langle b\rangle)$, and we must have $ab,a^{2}b\in\langle c\rangle$.  Suppose that $ab = c^m$ and $a^{2}b = c^n$, for non-zero integers $m$ and $n$.  Then $a^{-1}c^m = b = a^{-2}c^n$, so $a = c^{n-m}$, which contradicts $a\not\in\langle c\rangle$.