An Introduction to Theory of Groups by Joseph J. Rotman - Exercise $\textbf{(i)}$ Let $\sigma=(12345)$, let $P=\langle \sigma\rangle$ subgroup of $S_5$, and $N=N_{S_5}(P)$. Show that $|N|=20$ and $N=\langle\sigma, \alpha\rangle$ where $\alpha=(2354)$.
$\textbf{(ii)}$ If $A$ is the group (under composition), $$A=\{\phi:\Bbb Z_5\rightarrow \Bbb Z_5| \phi(x)=\alpha x+\beta, \alpha,\beta\in \Bbb Z_5, \alpha\neq 0\}$$
then $N\cong A$.(Hint:Show that $A=\langle s,t\rangle$, where $s:x\rightarrow x+1$, and $t:x\rightarrow 2x$)
My attempt is following.
Solution: 
$\textbf{(i)}$ $120=2^2.3.5$
Let $n_5$: the number of sylow $5$-subgroups. Then $n_5=6$.(Because $A_5$ only normal subgroups of $S_5$.)
Because $|S_5|=|N|[S_5:N]$ and $[S_5:N]=n_5$, so $|N|=20$.
Because of $(2354)(12345)(2354)^{-1} = (13524)$, then $(2354)=\alpha \in N$. So $\langle \alpha \rangle$ subgroup of $N$. Then
$$|\langle \alpha \rangle\langle \sigma \rangle |=\frac{|\langle \alpha \rangle| |\langle \sigma \rangle|}{|\langle \alpha \rangle \cap \langle \sigma \rangle|}=\frac{45}{1}=20$$ Hence, $N=\langle \alpha \rangle \langle \sigma \rangle=\langle \alpha, \sigma \rangle$.
For $\textbf{(i)}$.

  
*
  
*Is it correct? 
  
*I know  the product of any group can not be group. How can we see $\langle \alpha \rangle \langle \sigma \rangle (=\langle \alpha, \sigma \rangle)$ this product is group. To me it is group because of it is equal to group $\langle \alpha, \sigma \rangle$. Which property make them equal ? More general which propert make group of product of groups?
  
*I don't see why  $[G:N]=n_5$. I used just it. I remember this is related to $III$. Sylow Theorem. To me more complicated. How can I make easy to understand general concept and powers of normalizer?

For $\textbf{(ii)}$ I have some question.

- I don't know how can I start to solution.
- Why hint related to elements of group?
-How do we know when to use the elements of a group to find a group of isomorphisms, or when to try to construct a function?

 A: Part i) Your solution is correct. Note that you can just as easily count the number of $5$-cycles as follows. We have five slots: $(- - - - - )$, giving $5!$ ways to arrange $5$ letters. Now two permutations are equivalent if one can be obtained from the other by cyclic rotation, so we divide out by $5!$ yielding $4!$ distinct $5$-cycles. Now each $5$-cycle belongs to a $\mathbb{Z}_{5}$ subgroup of $\text{Sym}(5)$, which contains the identity element and four $5$-cycles. So we divide $4!/4 = 3!$ to obtain the number of Sylow-$5$ subgroups of $\text{Sym}(5)$. Being able to count in this way is helpful when dealing with larger groups, where less structural information may be immediately available.
Part ii) Whenever you have a group presentation for a group $G$, you can determine an isomorphism $\phi : G \to H$ be mapping the generators for $G$, and showing the image of the generators satisfies the same relations as $G$. This is often times much easier and less tedious than establishing an explicit bijection between two groups. I certainly don't want to tinker with every element of a group of order say $20$ to establish an isomorphism, when I could just as easily deal with a couple generators.
Now the hint is telling you to first show that:
$$A \cong \langle s, t \rangle$$
Where $s, t$ are given above. Can you show that $s, t$ generate $A$? Do this first. 
Once you have that $A \cong \langle s, t \rangle$, then you can establish an isomorphism $\phi : N \to A$ by mapping the generators of $N$ onto the generators of $A$.
