4
$\begingroup$

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.)

$\endgroup$
  • 2
    $\begingroup$ 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 $6$. $\endgroup$ – Tobias Kildetoft Dec 5 '13 at 14:35
3
$\begingroup$

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 $|G|=6$).

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I cannot use this theorem either $\endgroup$ – Gyt Dec 5 '13 at 15:31
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Rodney Coleman Dec 7 '13 at 17:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.