I have a lot of difficulty understanding this proof we went over in class about classification of groups of order 12.
Let $G$ be a finite group of order $n=p^rm$ where $m \nshortmid p$.
It seems that there is a mistake here one should read $p \nshortmid m$ or $gcd(p,m)=1$.
Denote $Syl_p(G)$ as the set of all sylow p-subgroups, and $n_p(G)$ as the cardinality of $Syl_p(G)$.
(1) G has subgroups of order $p^r$. (How do we know this??)
A priori, it has no reason to be true but this is a theorem known as $\textit{the first Sylow's theorem}$. It states that for any finite group $G$ and $p$ a prime number dividing $|G|$, if we write $|G|=p^rm$ with $p \nshortmid m$ then $G$ admits a $p$-Sylow (which is defined as a subgroup of $G$ whose cardinal is $p^r$).
(2) We know all sylow p-groups are conjugate and their number $n_p(G) | m$, by the 2nd and 3rd Sylow Theorems.
(3) We have that $n_p(G) \equiv 1(p). $
Then $n=12=2^2*3^1$ so $n_2(G)|3$ gives $n_2(G) \in \{1,3\}$ and $n_3(G)|4$ so $n_3(G) \in \{1,4\}$ since $2 \not\equiv 1(3)$.
Case 1: $n_3(G) = 4.$
Then the action of G by conjugation on $Syl_3(G)$ gives a homomorphism $f: G \rightarrow S_{Syl_3(G)} \simeq S_4$.
$Ker(f)$ consists of $g \in G$ that normalizes all 3-Sylow subgroups. (Why?).
How is $f$ defined ? Since it is given by the conjugation action $f(g):=S\mapsto gSg^{-1}$ so that if $f(g)$ is trivial (the indentity function here), it follows that $f(g)$ sends any $S$ to $S$ since $f(g)$ also sends any $S$ to $gSg^{-1}$ this boils down to for all $S$ we have $S=gSg^{-1}$.
Let $P_3$ be a 3-Sylow subgroup. Then the order of $P_3$ is 3. (I don't get that either)
This is the very definition of a $3$-Sylow in $G$ that gives you this since $|G|=12=4\times 3$.
, and $[G:N_G(P_3)] = 4.$ Thus $P_3 \subset N_G(P_3)$ gives $P_3 = N_G(P_3)$. (I'm lost here...)
Since $G$ acts transitively by conjugation (this is your point 2 : $p$-Sylow are all conjugate to each other) we have by the first class formula :
$$\frac{|G|}{n_3(G)}=|N_G(P_3)|\text{ whence } |N_G(P_3)|=3$$
By definition $H\subseteq N_G(H)$ so that $P_3\subseteq N_G(P_3)$ since both are of cardinal $3$ they are equal.
So $Ker(f) = \cap Syl_3(G) = \{e\}$ (the intersection of the sylow 3-subgroups is trivial. (How did we arrive here...)
We already saw that $Ker(f)$ is the intersection of normalizers of $3$-Sylows. Now we just saw that the normalizer of a $3$-Sylow is exactly the $3$-Sylow itself. Now I claim that if $S$ and $T$ are two different $3$-Sylows they cannot but intersect trivially. Indeed by Lagrange $S\cap T$ has cardinal dividing $|S|=3$. So the cardinal is either $3$ or $1$, if $|S\cap T|=3$ then $S\cap T\subseteq S$ and $|S|=3$ implies that $S\cap T=S$ and then that $S\subseteq T$ which implies (with the cardinal) that $S=T$ which is impossible. It follows that $|S\cap T|=1$ i.e. $S\cap T$ is trivial.
Since $Ker(f)$ is the intersection of $4$ different $3$-Sylows it is necessarily trivial.
Thus we conclude that $G$ is isomorphic to a subgroup of $S_4$ of order 12.
To show $G$ is isomorphic to $A_4$, we have that $G$ has 8 elements of order 3: 4 3-sylow subgroups each has 3-1=2 elements of order 3. And $S_4$ has 8 3-cycles meaning f(G) contains all 3-cycles and hence the group they generate, which is $A_4$, so $A_4 \subset f(G)$. But since $|A_4| = 12 = |f(G)$ then $f(G) = A_4$ and $G \simeq A_4$.
(Not understanding the first half of the proof, I do not get this part either).
Since $Ker(f)$ has been proven to be trivial it follows that $G$ is isomorphic to $f(G)$ which is a subgroup of $S_4$. Now you would like to identify the subgroup $f(G)$ you know that it contains $4$ $3$-Sylows. A $3$-Sylow is of order $3$, it contains two elements of order $3$ and one of order $1$. Since the intersection of two different $3$-Sylows is trivial and since you have $4$ $3$-Sylows you finally get in $f(G)$ exactly $8$ elements of order $3$ in $f(G)$. Since they are elements of order $3$ in $S_4$ they cannot but be $3$-cycles and since there are $8$ of them in $S_4$ they are all contained in $f(G)$. The fact that they generate $A_4$ is classical so that $A_4\leq f(G)$ and by cardinality argument they are equal.
This part is too complicated, one way to define $A_n$ is to say that this is the unique subgroup of index $2$ in $S_4$. Since $S_4$ is of order $24$ and $f(G)$ of order $12$ it follows that $f(G)$ is a subgroup of index $2$ in $S_4$ and then $A_4=f(G)$ by unicity.
Case 2: $n_3(G) = 1.$ The scope of the question is too large, I will need to post separately to understand the second half, unless the organizational rules of math.stackexchange would require that I post here. In that case, I'll edit the question.
I basically cannot follow the proof because it seems to skip too many steps; it would be helpful if anyone could elaborate case 1 with more reasoning so that I can follow it. Any help would be appreciated; I've spent hours trying to look up and decipher the proof but without any success.
Best Answer
Note that Sylow theorems give you necessary conditions for the number of Sylow $p$-groups. Those conditions are not sufficient. In your case, you have proven that $|Syl_5(G)|=1$ and $|Syl_3(G)|\in\{1,25\}$. This means that $|Syl_3(G)|$ and $|Syl_5(G)|$ cannot take any other values than those you've noted - but it doesn't mean they must take all of them.
(The case where those numbers are all $1$ is a bit of an exception. There is always a group where all the numbers of Sylow $p$-groups are $1$. Indeed - take a direct sum of the group's Sylow $p$-groups!)
Which leaves us with the case $|Syl_5(G)|=1$ and $|Syl_3(G)|=25$, for which we still don't know whether it is possible or impossible. Suppose that it is possible, and let's try to either derive a contradiction or construct the group $G$ in which this is true. What we do know is that the Sylow $5$-group is unique (call it $H$, $|H|=25$) and is therefore normal in $G$. On the other hand, you can pick a Sylow $3$-subgroup $K$, $|K|=3$, which is most certainly not normal (Sylow's 2nd theorem - Sylow $3$-groups are all conjugates of each other, so $K$ has $25$ conjugates). We also know that:
Now, the map $\theta_a:h\mapsto a^{-1}ha$ is an automorphism of $H$ and its order must divide the order of $a$ - so its order is either $1$ or $3$. If its order is $1$, however, it means that $a^{-1}ha=h$, i.e. $ah=ha$ for all $h\in H$. This means that every element of $H$ commutes with every element of $K$, and so $G$ turns up a direct sum of $H$ and $K$ - and so $Syl_3(G)=1$.
Thus, the question here really boils down to: is there an automorphism of order $3$ of either $C_{25}$ or $C_5\times C_5$?
Let's check the two cases:
Let us just show how this automorphism acts on $C_5\times C_5$: if the elements of $C_5\times C_5$ are represented as $u^iv^j$ where $u$ and $v$ are the generators of the two $C_5$'s and $i,j\in\mathbb Z_5$, then $\theta_A(u^iv^j)=u^{-i+j}v^{-i}$, because $\begin{bmatrix}-1&1\\-1&0\end{bmatrix}\begin{bmatrix}i\\j\end{bmatrix}=\begin{bmatrix}-i+j\\-i\end{bmatrix}$.
With that automorphism, your group $G$ can be constructed, as shown above, as $(C_5\times C_5)\rtimes_{\theta_A}C_3$, and it will have $25$ Sylow $3$-groups.