TITLE: Show that every group $G$ of order $175$ is abelian and list all isomorphism types of these groups
QUESTION [5 upvotes]: Show that every group $G$ of order 175 is abelian and list all isomorphism types of these groups. [HINT: Look at Sylow $p$-subgroups and use the fact that every group of order $p^2$ for a prime number $p$ is abelian.]

What I did was this. $|G| = 175$. Splitting 175 gives us $175 = 25 \cdot 7$. Now we want to calculate the Sylow $p$-groups, i.e we want
$$P= n_7: n_7 \equiv 1 mod 7 \hspace{1.5cm} n_7|25$$
$$Q= n_{25}: n_{25} \equiv 1 mod 25 \hspace{1.5cm} n_{25} | 7$$
After listing all elements that are $\equiv 1 mod 7$ and $1 mod 25$ you see that the only (avaliable) ones are $n_7 = n_{25} = 1$. This tells us that both groups $P,Q$ are normal subgroups of $G$. I think, by definition of a normal subgroup, they are abelian and so this tells us that $G$ is abelian. To list all the isomorphism types, we want the semidirect product (SDP) such that
$$P \rightarrow Aut(Q) = C_7 \rightarrow C_{20}$$
As there are no elements of order 7 in $C_{20}$, the only SDP we have is the trivial SDP, i.e the direct product
$$C_7 \times C_{25} \cong C_{175}$$
We know that $175 = 5^2 \cdot 7$ and so multiplying the powers shows us that there are 2 non-isomorphic groups:
$$C_{25} \times C_7$$
$$C_5 \times C_5 \times C_7$$
My question for this is is my reasoning also correct for things like showing the abelian groups? I saw something which said something about $P \cap Q = I_G$ and they used this but I don't understand what it was.
The next question, assuming that I had to possibilites for my $p$ subgroup, i.e $n_p = 1 or x$, how would I go about answering this question? (I am doing a question like this now and am stuck as I have two Sylow $p$-subgroups).

REPLY [4 votes]: The fact that $P, Q $ are both normal tells you that $G$ is a direct product of $P$ and $Q$. $P$ is abelian because of the hint given in the problem statement: $|P| = 25 = 5^2$, and $5$ is prime.$|Q| = 7$ with $7$ prime. All groups of prime order are cyclic, and all cyclic groups are abelian, $Q$ is therefore abelian. 
Hence $G = P\times Q$, as the direct product of two abelian groups, is therefore abelian.

Just to confirm/answer your first question (listing all possible abelian groups of order 175):
Yes, you correctly showed that if a group $G$ of order 175 is abelian, then it is isomorphic to exactly one of two (non-isomorphic) groups:
$$G \cong C_{175} \cong C_{25} \times C_7$$
$\quad\quad\quad\quad\text{**OR**}$
$$G\cong C_5 \times C_5 \times C_7$$
by the Fundamental Theorem of Finitely Generated Abelian Groups.

Note (added given comment/question below): 
$C_{mn} \cong C_m\times C_n$ if and only if $\gcd(m, n) = 1$.
So we have that $C_{175} \cong C_{25}\times C_7$ since $\gcd(25, 7) = 1$.
On the other hand, $C_{25} \times C_7 \ncong (C_5 \times C_5) \times C_7$ since $\gcd(5, 5) \neq 1$