# Finding 2 nonabelian, nonisomorphic groups of order 225

I need to find 2 nonabelian, nonisomorphic groups of order 225. Here's what I have so far:

Let $G$ be a group of order 225.

By Sylow's theorems, we have that $G$ contains $P_{25}$ and $P_{9}$, subgroups of order $25$ and $9$, respectively. It's easy to see that $n_5 = 1$ and, so, $P_{25}$ is normal in $G$. Therefore, $G = P_{25}P_9$, since $P_{25} \cap P_9 = \{1\}$.

Therefore, $G \simeq P_{25} \rtimes_{\varphi} P_9$ for some $\varphi: P_9 \rightarrow \text{Aut}(P_{25})$. Since $P_{25}$ and $P_9$ are both abelian, we need $\varphi$ to be non-trivial.

Now, $P_{25}$ cannot be cyclic (because that would make $G$ abelian), and so, $P_{25} = \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}$. We know that $\text{Aut}(\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}) = GL(2,5)$ which has order $480$, and so, $|\varphi(P_9)| = 1$ or $3$ and, since we need $\varphi$ non trivial, we have $|\varphi(P_9)| = 3$.

Now, we have two choices (upto isomorphism) for $P_9$, $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ and $\mathbb{Z}/9\mathbb{Z}$. I figure finding such a $\varphi$ for each of these choices will give me what I need.

However, I get stuck here. I tried to find an element of order $3$ in $GL(2,5)$ (for the case where $P_9$ is the cyclic group), but I'm not sure how to do this easily. I'm not really sure how to proceed when $P_9 = \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ either.

• Suppose you find an element s of order 3. In the first $P_9$ case, you can send the generator to $s$. In the second, you can send $(1, 0)$ to $s$. Finding that element may take a little cleverness, though. – John Hughes Nov 12 '17 at 1:26
• Finding that element is exactly where I am stuck haha – jackson5 Nov 12 '17 at 1:31
• The matrix $\begin{pmatrix} -1&-1\\1&0\end{pmatrix}$ has order 3 over any field. – Dane Nov 12 '17 at 1:38
• @stressed-out oops, I meant $P_{25}$ and $P_9$ are both abelian, since they have order of the form $p^2, p$ prime. If $P_{25}$ is cyclic, it forces $\varphi$ to be trivial (since $|\text{Aut}(C_{25})| = 20$ and $\gcd(9,20) = 1$), which forces $G$ to be abelian. – jackson5 Nov 12 '17 at 1:41
• Well...you DID say "I'm not sure how to proceed when $P_9 = ...$ either." I was addressing that part, not the "find an element of order 3". – John Hughes Nov 12 '17 at 3:08