Classifying Groups of Order 28 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!
 A: 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.
A: The three maps $\psi_{2,3,4}$ differ by an automorphism (or, if you want, coordinate change) on $K$: $\psi_3$ is obtained from $\psi_2$ by first applying the map $K\to K$, switching $a$ and $b$; Similarly the map that maps $a\mapsto a$,$b\mapsto ab$ prepended to $\psi_2$ yields $\psi_4$.
This gives you the isomorphisms -- e.g. if the product using $\psi_2$ is generated by $a_2,b_2,h_2$ (the latter being a generator of the 7-Sylow subgroup, and ditto with index $3$ for the product using $\psi_3$, then the isomorphism is $a_2\mapsto b_3$, $b_2\mapsto a_3$, $h_2\mapsto h_3$.
A: The three maps $\psi_2$, $\psi_3$ and $\psi_4$ differ by automorphisms of $K$. That is to say, there exist $\varphi_{ij}\in\operatorname{Aut}{K}$ such that $\psi_j=\psi_i\circ\varphi_{ij}$, for all $i,j\in\{2,3,4\}$. For example, for $i=2$ and $j=3$ we may take the automorphism of $K$ defined by
$$\varphi_{ij}:\left\lbrace \begin{array}{c}
a\mapsto b\\
b\mapsto a
\end{array}\right\rbrace.$$
It is not difficult to find similar automorphisms for the other choices of $i$ and $j$.
Then it is a matter of routine verification to show that the map
$$\Phi_{ij}:\ H\rtimes_{\psi_i}K\ \longrightarrow\ H\rtimes_{\psi_j}K:\ \big(h,k\big)\ \longmapsto\ \big(h,\varphi_{ij}(k)\big),$$
is a group isomorphism: For any two pairs $(h_1,k_1),(h_2,k_2)\in H\rtimes_{\psi_j}K$ we have, by definition
$$(h_1,k_1)\cdot_{\psi_j}(h_2,k_2)=(h_1\psi_j(k_2)(h_2),k_1k_2)$$
and so to show that
$$\Phi_{ij}\big((h_1,k_1)\cdot_{\psi_i}(h_2,k_2)\big)=\Phi_{ij}\big(h_1,k_1\big)\cdot_{\psi_j}\Phi_{ij}\big(h_2,k_2\big),$$
we simply expand both sides:
\begin{eqnarray*}
\Phi_{ij}\big((h_1,k_1)\cdot_{\psi_i}(h_2,k_2)\big)
&=&\Phi_{ij}\big(h_1\psi_i(k_2)(h_2),k_1k_2\big)\\
&=&\big(h_1\psi_i(k_2)(h_2),\varphi_{ij}(k_1k_2)\big)\\
&&\\
\Phi_{ij}\big(h_1,k_1\big)\cdot_{\psi_j}\Phi_{ij}\big(h_2,k_2\big)
&=&\big(h_1,(\psi_j^{-1}\circ\psi_i)(k_1)\big)\cdot_{\psi_j}\big(h_2,\varphi_{ij}(k_2)\big)\\
&=&\big(h_1\psi_j\big(\varphi_{ij}(k_2)\big),\varphi_{ij}(k_1)\varphi_{ij}(k_2)\big)\\
&=&\big(h_1\psi_i(k_2),\varphi_{ij}(k_1k_2)\big)\\
\end{eqnarray*}
This shows that the three semidirect products are indeed isomorphic.

Of course the proof above is rather general; for any group homomorphism
$$\psi:\ K\ \longrightarrow\ \operatorname{Aut}(H),$$
and any automorphism $\varphi\in\operatorname{Aut}(K)$ there is this exact same isomorphism
$$H\rtimes_{\psi}K\cong H\rtimes_{\psi\circ\varphi}K.$$
