how did we know the action in $\mathbb{Z}_{11} \rtimes_{\varphi} \mathbb{Z}_{5}$?

Question

our professor was explaining to us $\mathbb{Z}_{11} \rtimes_{\varphi} \mathbb{Z}_{5}$ and he gave us the following automorphism:

$\varphi: \mathbb{Z}_{11} \rightarrow \mathbb{Z}_{11} $ defined by $x \mapsto ux$ where $u \in \mathbb{Z}_{11}^{*}$

Then he said $Aut(\mathbb{Z}_{11}) \cong \mathbb{Z}_{10}$ and added that $\mathbb{Z}_{10}$ has a unique subgroup $H \cong \mathbb{Z}_{5}$ and he added so there is a $1-1$ map $\varphi$ from $\mathbb{Z}_{5}$ to $Aut(\mathbb{Z}_{11})$ so $\mathbb{Z}_{5}$ acts on $\mathbb{Z}_{11} .$

My question is:

1-I do not understand why my professor took the $\varphi$ like that? could anyone explain that for me please?

2-I do not understand why all the previous leads us that $\mathbb{Z}_{5}$ acts on $\mathbb{Z}_{11}$, the logic in his conclusion is not clear for me, could anyone explain that for me please?

3- In general, how can we know the action of the second group on the first group?

Note:

The ingredients needed to make a semi-direct product are two groups H,K and a group action of H (the second group) on K (the first group). Let ${}^hg$ denote the action of h in H on k in K. Then the semi-direct product is the Cartesian product KxH (as a set) with multiplication defined by $(k,h)(k'h')=(k({}^hk'),hh').$

If the action is trivial (that is, ${}^hk = k$ for all h,k), then the semi-direct product is the more familiar direct product.

Answer 1

First, there is a certain amount of reading and experimenting that you are going to have to do in order to know this subject better.

Now, to answer one of your questions, a general result is that $\rm{Aut}\Bbb Z_n\cong\Bbb Z_n^×$, where the latter is the group of units. When $n=p$ is prime, it turns out that the group of units is cyclic of order $p-1$. That's $\Bbb Z_p^×\cong\Bbb Z_{p-1}$.

Thus, in your case, we have $\rm{Aut}\Bbb Z_{11}\cong\Bbb Z_{10}$.

Next a cyclic group has a unique subgroup of any order dividing the order of the group.

Finally, for a semi-direct product, we require a homomorphism from, in this case, $\Bbb Z_5$ to $\rm{Aut}\Bbb Z_{11}\cong\Bbb Z_{10}$. Since $\Bbb Z_{10}$ has a subgroup isomorphic to $\Bbb Z_5$, we do have a nontrivial homomorphism. This can also be described as a group action, by the way.