If a non-abelian group G of order 12 has a normal 3-subgroup, and a 4-subgroup that is not cyclic, then G contains an element of order 6

Question

I have tried to count the number of elements by their order. Since G is not abelian, there is no element of order 12. There is one element of order 1, 2 elements of order 3, and no elements of order 4 since one 2-sylow not being cyclic implies that non of the 2-sylows are cyclic. Now I only have to show that the number of elements of order 2 is less than 9 but I don't know how to.

Thanks in advance for your help!

Answer 1

$G$ contains an element of order $2$ say $b$.

$G$ contains an element of order $3$ say $a$ where $\langle a\rangle =\{e,a,a^2\}$ is a normal Sylow-$3$ subgroup of $G$.

Now $bab^{-1}\in \langle a\rangle\implies bab^{-1}=a$ or $bab^{-1}=e$ or $bab^{-1}=a^2$

If $bab^{-1}=a\implies ab=ba\implies o(ab)=o(a)o(b)=6$

If $bab^{-1}=a^2\implies ba=a^2b$. Now consider the subgroup $\langle a,b\rangle\cong S_3$ which is a subgroup of $G$ .

Also $G$ has $K_4=\{e,a,b,c\}$ the Kleins group as its subgroup .

Then take any $3$ cycle from $S_3$ say $(132)$ and $a\in K_4$. Then $((132),a)$ has order $6$