Classifying all non-abelian groups of order $55$ 
Suppose a group $G$ is the semidirect product of normal subgroup $N$ and subgroup $H$, i.e., $G=N\rtimes_\varphi H$. Find all semidirect products (up to isomorphism) of $N=\mathbb Z_{11}, H=\mathbb Z_5.$


Can someone help me solve this question? I have spent a long time in this question, but still not sure how to analyze this question.
 A: Suppose $G= \mathbb Z_{11}\rtimes_\varphi \mathbb Z_5$ where $\mathbb Z_{11}$ and $\mathbb Z_5$ denote the cyclic groups of order $11$ and $5$ respectively, and $\varphi: \mathbb Z_5\to\text{Aut}(\mathbb Z_{11})$ is a homomorphism from $\mathbb Z_5$ to the automorphim group of $\mathbb Z_{11}$. So now it suffices to classify $\varphi $. First, we prove a lemma.

Lemma $\ $  Suppose $p$ is prime, then the automorphism group of $(\mathbb Z_{p},+,0)$ is $(\mathbb Z_{p}^*,\times,1)\cong (\mathbb Z_{p-1},+,0)$.
Proof: $\ $ For each $r\in\mathbb Z_p^*$, and one can easily check that
\begin{align} f_r:\mathbb Z_p&\to \mathbb Z_p\\ a&\mapsto ra \end{align} determines an automorphism of $\mathbb Z_p$ and they are all the automorphisms. Now consider the map:
\begin{align}
\phi: \mathbb Z_{p}^* &\to \text{Aut}(\mathbb Z_p)\\
r&\mapsto f_r
\end{align}
which is apparantly a bijection. It leaves us to check that $\phi$ is a group homomorphism. Note that for every $a\in\mathbb Z_p$:
\begin{align}
\phi(r_1)\phi(r_2)(a)=f_{r_1}(f_{r_2}(a))=r_1r_2a=f_{r_1r_2}(a)=\phi(r_1r_2)(a)
\end{align} and we are done.

Now in our case, $\text{Aut}(\mathbb Z_{11})\cong \mathbb Z_{10}$. We consider the group homomorphisms: $\varphi: \mathbb Z_5\to\mathbb Z_{10}$ which are uniquely determined by $\varphi (1)$.
If $\varphi(1)=0$ then we have the trivial homomorphism, otherwise, we have:

*

*$\varphi_1(1)=2$ corresponding to $f_4$;


*$\varphi_2(1)=4$ corresponding to $f_5$;


*$\varphi_3(1)=6$ corresponding to $f_9$;


*$\varphi_4(1)=8$ corresponding to $f_3$.
That is because $2$ happens to be a generator of $\mathbb Z_{11}^*$ and $2^2=4, 2^4=5, 2^6=9, 2^8=3$ in the sense of modulo $11$, and the isomorphism between $\mathbb Z_{11}^*$ and $\mathbb Z_{10}$ we construct by
\begin{align}
\displaystyle \mathbb Z_{10}&\to\text{Aut}(\mathbb Z_{11})\\
r&\mapsto f_{2^r}
\end{align}
However, we claim that $\mathbb Z_{11}\rtimes_{\varphi_1}\mathbb Z_5\cong\mathbb Z_{11}\rtimes_{\varphi_2}\mathbb Z_5\cong\mathbb Z_{11}\rtimes_{\varphi_3}\mathbb Z_5\cong\mathbb Z_{11}\rtimes_{\varphi_4}\mathbb Z_5$. Now I leave you to check the following isomorphisms:
\begin{align}
\mathbb Z_{11}\rtimes_{\varphi_1}\mathbb Z_5&\to\mathbb Z_{11}\rtimes_{\varphi_2}\mathbb Z_5\\
(n,h)&\mapsto (3n,h)
\end{align}
\begin{align}
\mathbb Z_{11}\rtimes_{\varphi_1}\mathbb Z_5&\to\mathbb Z_{11}\rtimes_{\varphi_3}\mathbb Z_5\\
(n,h)&\mapsto (9n,h)
\end{align}
\begin{align}
\mathbb Z_{11}\rtimes_{\varphi_1}\mathbb Z_5&\to\mathbb Z_{11}\rtimes_{\varphi_4}\mathbb Z_5\\
(n,h)&\mapsto (5n,h)
\end{align}
Therefore, there are exactly one non-abelian group $\mathbb Z_{11}\rtimes\mathbb Z_5$ of order $55$ up to isomorphism.
