I am trying to classify groups of order 28. In the course of the problem, I am stuck in showing that three semidirect products are isomorphic to each other. In this problem, $G$ is a group of order 28, $H\in\mathrm{Syl}_{7}(G)$ is the unique Sylow 7-subgroup, and $K\in\mathrm{Syl}_{2}(G)$. I am working on the case where $K\cong \mathbb{Z}_{2}\times \mathbb{Z}_{2}$.
We have the following groups to consider: $$K\cong\mathbb{Z}_{2}\times \mathbb{Z}_{2}=\left\langle a,b\:|\:a^{2}=b^{2}=(ab)^{2}=1\right\rangle$$ $$\mathrm{Aut}(H)\cong\mathbb{Z}_{6}=\left\langle x\:|\: x^{6}=1\right\rangle$$
Let $\psi_{j}: K\to \mathrm{Aut}(H)$, with $j\in\{1,2,3,4\}$, be defined as follows:
$$\psi_{1}:\left\lbrace \begin{array}{c}
a\mapsto 1\\
b\mapsto 1
\end{array}\right\rbrace \:\:\:\:\:\psi_{2}:\left\lbrace \begin{array}{c}
a\mapsto x^{3}\\
b\mapsto 1
\end{array}\right\rbrace\:\:\:\:\:\psi_{3}:\left\lbrace \begin{array}{c}
a\mapsto 1\\
b\mapsto x^{3}
\end{array}\right\rbrace\:\:\:\:\:\psi_{4}:\left\lbrace \begin{array}{c}
a\mapsto x^{3}\\
b\mapsto x^{3}
\end{array}\right\rbrace$$
I know that because $\psi_{1}$ is trivial, we get $H\rtimes_{\psi_{1}}K\cong H\times K$. With all the previous work that I have done for this problem, this direct product determines the third isomorphism class for my isomorphism types. The problem statement tells me that there are four isomorphism types, so I only need one more. This means that we need
$$H\rtimes_{\psi_{2}}K\cong H\rtimes_{\psi_{3}}K\cong H\rtimes_{\psi_{4}}K.$$
However, I do not know how to show that all these semidirect products are actually isomorphic. Thanks in advance for any help!
Best Answer
The other answer is definitely correct. Here is a different way to think about it. In each case, one of the elements of $C_2\times C_2$ acts trivially on $C_7$. In the first case, this is $b$. So, in $G$, consider $\langle b\rangle$. It commutes with $x$, and commutes with $a$. But in all cases, $G=\langle x,a,b\rangle$. Thus $b$ is central, and $\langle x,a\rangle$ is a normal subgroup as well. Thus $G$ is the direct product of $C_7\rtimes C_2$ and $C_2$. Since you have (probably) shown before that there is a unique non-abelian group $C_7\rtimes C_2$, we obtain that $G$ is simply $D_{14}\times C_2$.
But the same is true in the second case, with the roles of $a$ and $b$ swapped. Thus these two groups are both $D_{14}\times C_2$, and so are isomorphic.
Finally, the third case. Now we have $G\cong \langle x,a\rangle\times \langle ab\rangle$, and the proof goes through again. Thus all three groups are isomorphic, because they are the same direct product.
Having said that, Alexander Hulpke's answer is the correct one, because it is more generally useful than an in-depth look at the groups.