Q:

Classifying groups of order $5 \cdot 11 \cdot 61$

My question is whether I'm classifying groups of order $5 \cdot 11 \cdot 61$ correctly.  (This is a qualifying exam question, so I also want to make sure that I'm doing it “efficiently”.)
Sylow's theorem shows that
$$n_{11} = n_{61} = 1\qquad \textsf{ and }\qquad n_5 = 1, \,11,\, 61,\, 11 \cdot 61.$$
Let $P = \mathrm{Syl}_{5} = \langle x \rangle$ (chosen arbitrarily), $Q = \mathrm{Syl}_{11} = \langle y \rangle$, and $R = \mathrm{Syl}_{61} = \langle z \rangle$.  Since $Q$ and $R$ are normal and intersect trivially, $yz = zy$.  Moreover, since $P$ intersects trivially with $Q$ and $R$,
$$PQ \cong Q \rtimes_\gamma P \qquad \textsf{ and }\qquad G = PQR \cong R \rtimes_\varphi PQ,$$
where $\gamma \colon P \to \mathrm{Aut}(Q) \cong \mathbf{Z}_{10}$ and $\varphi \colon PQ \to \mathrm{Aut}(R) \cong \mathbf{Z}_{60}$ are the usual conjugation homomorphisms.  
$\gamma$ is determined by where $x$ is sent.  We have $|\gamma_x| \big| 5 \text{ and } 10$, so $|\gamma_x| = 1 \text{ or } 5$.  If $\gamma_x$ is trivial, we obtain the relation $xy = yx$.  If $\gamma_x$ is not trivial, then it is a generator of $\mathbf{Z}_5 \subset \mathbf{Z}_{10}$; moreover, the resulting semidirect product is independent of which generator $x$ is sent to.  One can check quickly that the automorphism $y \mapsto y^2$ generates $\mathbf{Z}_{10}$.  So, possibly after rechoosing $y$, we may take $\gamma_x \colon y \mapsto y^4$ and we obtain the relation $y^4 = xyx^4$.  
$\varphi$ is determined by where $x$ and $y$ are sent.  We have $|\varphi_x| \big| 5 \text{ and } 60$ and $|\varphi_y| \big| 11 \text{ and } 60$, so $|\varphi_x| = 1 \text{ or } 5$ and $|\varphi_y| = 1$ (which we already knew since $zy = yz$).  If $\varphi_x$ is trivial, then we obtain the relation $xz = zx$.  If $\varphi_x$ is not trivial, then let $1 < N < 61$ be chosen such that the automorphism $z \mapsto z^N$ generates $\mathbf{Z}_{60}$ (I gather that there is no general formula for $N$).  Then, possibly after rechoosing $z$, we may take $\varphi_x \colon z \mapsto z^{6N}$ and we obtain the relation $z^{6N} = xzx^4$.  
We conclude 
$$
G  \cong \left\langle x, y, z : \begin{array}{cc}
x^5 = y^{11} = z^{61} = 1, & yz = zy,\\[0.05in]
\left[ xy = yx \;\textsf{ or }\; y^4 = xyx^4 \right], & \left[ xz = zx \;\textsf{ or }\; z^{6N} = xzx^4 \right]
\end{array}\right\rangle$$
I suppose to be fully rigorous I would have to prove that there exists such a group for each of the four combinations and that they are all distinct...
Any comments, corrections, answers, etc., are warmly welcomed.  

A:

This is not an easy question and you have made a subtle mistake. If $\gamma_x$ and $\phi_x$ are both trivial then you get an abelian group. If one of them is trivial and the other is not, then you are correct in saying that, up to group isomorphism, there is a unique choice for the nontrivial one. So we have three isomorphism types of groups so far.
The remaining case is when $\gamma_x$ and $\varphi_x$ are both nontrivial. In order to get your preferred choice of $\gamma_x$ you may have to replace $x$ by a power. You cannot then replace $x$ by a power again to get your preferred choice for $\varphi_x$. So, having fixed $x$ and $\gamma_x$, the four possible choices of $\varphi_x$ in give rise to four nonisomorphic groups.
So there are actually $7$ isomorphism types of groups of this order. 

