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I am trying to classify all groups of order $3825=3^2 \cdot 5^2 \cdot 17$. The Sylow theorems indicate that the number of Sylow p-subgroups for each p rime are $n_{17}=1$, and $n_{3}=1,25,85$ and $n_5=1,51$. Moreover, I know that the Sylow $3-$ and $5-$ subgroups $P_3$ and $P_5$ are abelian, as they are of the form $|G|=p^2$. I have classified all abelian groups using the structure theorem/fundamental theorem.

How do I find out the non-abelian groups of this order? In particular how do I realise $G$ as a semi-direct product with $P_{17}$ as one of the factors, as done while classifying groups? As $P_3$ and $P_5$ are not known to be normal, I cannot form something like $P_3 P_5$ as a subgroup. Counting arguments are not strong enough and don't tell me anything about $n_5$ and $n_3$.

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First prove that Sylow $17$- subgroup is central (using N-C theorem). Then this will imply $n_3=25$, $n_5=1$ since $17 | N_G(S_3)$ and $17 | N_G(S_5)$, where $S_p$ denotes Sylow $p$-subgroup. Since $S_5$ and $S_17$ are normal, we can make a group $H$ generated by these two subgroups and $|H|=5^2 *17$ .Now $G$ is semi-direct product of $H$ with $S_3$.Thus, there exists a homomorphism $\phi$ from $S_3$ to $Aut(H)$. By divisibility conditions, $ker$ $\phi = S_3$ or $ker$ $\phi = Z_3$. Therefore if $ker$ $\phi = S_3$, $S_3$ centralizes $H$. Since $S_3$ is abelian, $S_3$ is central, contradicting $n_3=25$.Therefore, all such $G$ are abelian. If $ker$ $\phi = Z_3$, and $S_5=C_5^2$ then $G$ is a semi-direct product of a unique subgroup $T=Hker\phi$ of order $3*5^2 *17$ with $Z_3$.

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This answer concerns only non abelian groups $G$ only, the abelian case can be easily derived. From the Sylow theorems it follows that $n_{17} = 1$ so there exists a normal subgroup $N$ of order $17$ of $G$. Let's see which groups $H = G/N$ of order $225$ we can construct. For those abelian ones none of their automorphism groups contains a factor $17$ so the only groups $G$ that can be made out of them are direct products with $N$. What non-abelian groups can be made of order $225$. Using the Sylow theorems again we find that $H$ has a normal subgroup $M$ of order $25$. Whe have either $M = C_5 \times C_5$ or $M = C_{25}$. In the latter case $\operatorname{Aut}(M)$ has order $20$ so only a direct product group is possible, but in the former case we have $\operatorname{Aut}(M) = \operatorname{GL}(2,5)$ which contains elements of order $3$ that are all conjugate so this gives us two possibilties for $H$ being $(C_5 \times C_5) \rtimes C_9$ and $((C_5 \times C_5) \rtimes C_3)\times C_3$, so there are up to isomophism two non abelian groups of order 3825.

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