# Existence of a subgroup with order 3 in a group with order 6

Let $G$ be a group of order 6. Why does $G$ has a subgroup of order 3 even if $G$ isn't cyclic?

I've tried using to use negation and assume all elements in $G$ have an order of 2 or 1 but I can't contradict with that (without sylow or cauchy theorem.)

• If all elements have order $2$ then the group is abelian. Quotient by a subgroup of order $2$ to get element or order $3$ or $6$. – Tobias Kildetoft Dec 5 '13 at 14:35

## 3 Answers

If every non-identity element in $G$ has order $2$, then $G$ is abelian [Proof: $gh=(gh)^{-1}=h^{-1}g^{-1}=hg$.].

If $a$ and $b$ are two elements of order $2$ in an abelian group $G$, then $\langle a,b\rangle = \{1,a,b,ab\}$ is a subgroup of order $4$, violating Lagrange's Theorem (since $|G|=6$).

If all elements have order two, then, as @Tobias points out, the group is abelian ($a b (b a)^{-1} = ab ab = (ab)^2 = 1.$ An abelian group is a product of cyclic groups, and since all elements have order $2,$ the order must be a power of $2,$ which $6$ is not. Of course, only you know if you are allowed to use the structure theorem for abelian groups.

• I cannot use this theorem either – Gyt Dec 5 '13 at 15:31

The group G has a Sylow 3-subgroup H which is of order 3. As H is cyclic there is an element of G of order 3.

• This can be generalized. If a group has order pm, where p is prime and does not divide m, then G has an element of order p. – Rodney Coleman Dec 7 '13 at 17:14