# Group of Order 44.

Question. Let $G$ be a group of order $44$ such that it has a subgroup isomorphic to $\mathbb Z/2\oplus \mathbb Z/2$ and a subgroup isomorphic to $\mathbb Z/4$.

Show that $G$ does not exist.

Attempt. Let $n_{11}(G)$ be the number of Sylow-$11$ subgroups of $G$. Then using the third Sylow theorem we have $n_{11}(G)\equiv 1\pmod{11}$, giving $n_{11}(G)=1$. Now by Sylow Conjugation we conclude that a subgroup of order $11$ in $G$ is normal.

Let $H$ be a subgroup of order $11$ in $G$ and assume on the contrary that $K_1,K_2<G$ with $K_1\cong \mathbb Z/2\oplus \mathbb Z/2$ and $K_2\cong \mathbb Z/4$.

We know that, since $H$ is normal in $G$, that $HK_1,HK_2$ are subgroups of $G$. It is also easy to see that $HK_1=HK_2=G$. But here I am stuck.

• Use one of the other Sylow theorems (forgot the number), all $p$-Sylow subgroups are conjugate. – Daniel Fischer May 9 '14 at 19:49
• @DanielFischer Ah! Thanks. – caffeinemachine May 9 '14 at 20:07

If $P,Q\in \operatorname{Syl}_p(G)$ there is an (inner) automorphism $\varphi\in\operatorname{Aut}(G)$ such that $\varphi(P)=Q$. Then $\varphi\!\mid_P:P\rightarrow Q$ is an isomorphism. In particular, all Sylow $p$-subgroups must be isomorphic. In this case $4$ is the highest power of $2$ dividing $|G|$, so we cannot have Sylow $2$-subgroups isomorphic to both $V$ and $C_4$.