# 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.

• I hope you don't mind my edits, I got maybe a bit carried away - feel free to undo anything you don't like. – Zev Chonoles Jun 27 '15 at 4:30
• Edits are great, thanks. – Doug Jun 27 '15 at 4:32

## 2 Answers

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.

• To see where I went wrong in my solution, I did recognize the problem you brought up, and I thought I had corrected for it (note that above I was adjusting my choices of $y$ and $z$, not $x$), but my "correction" was incorrect. To see why, suppose it happens that $\gamma_x \colon y \mapsto y^4$. Then you cannot choose another generator $y^\prime$ such that $\gamma_x \colon y^\prime \mapsto \left( y^\prime \right)^2$. (continued) – Doug Jun 28 '15 at 19:42
• Indeed, suppose you could. Write $y^\prime = y^\alpha$. Then $\gamma_x \colon y^\prime \mapsto y^{2\alpha} = y^{4\alpha}$. Hence $\left( y^\prime \right)^2 = \left( y^\alpha \right)^2 = 1$, but the order of $y^\prime$ is 11, hence we obtain a contradiction. – Doug Jun 28 '15 at 19:46
• And, again, to be fully rigorous, one would have to show that the resulting groups are mutually non-isomorphic and actually exist (for the latter, I believe you can just use semidirect products to construct them "abstractly" out of cyclic groups). Hopefully, on a qualifying exam, one could omit this last step without being penalized too harshly (if at all). – Doug Jun 28 '15 at 19:49
• @Doug As you say, the existence of the groups folows from the general theory of semidirect prodcuts, which you could probably assume. You should be able to prove without too much trouble (by considering the orders of elements in the groups for example) that the first three groups are mutually non-isomorphic and not isomorphic to any of the final four, which look more alike. Proving them non-isomorphic is a bit harder. As a first step you could show that, if two of them are isomorphic, then there must be an isomorphicm mapping $x,y,z$ in the first group to powers of $x,y,z$ in the second group. – Derek Holt Jun 28 '15 at 19:59

I am going to start where Derek finished, with the aim of giving some justifications as to why the groups are mutually non-isomorphic.

To summarize, there are seven possibilities: we may choose $x, y, z$ such that $P = \langle x \rangle \cong \mathbf{Z}_5$, $Q = \langle y \rangle \cong \mathbf{Z}_{11}$, $R = \langle z \rangle \cong \mathbf{Z}_{61}$, and $G = PQR$, where $yz = zy$, and one of \begin{cases} xy = yx \\ xz = zx \end{cases} or \begin{cases} xy = yx \\ xzx^{-1} = z^N \end{cases} or \begin{cases} xyx^{-1} = y^2 \\ xz = zx \end{cases} or \begin{cases} xyx^{-1} = y^{2\alpha}, & \alpha = 1, \dots, 4 \\ xzx^{-1} = z^N. \end{cases}

The most trivial case is the first, where $G \cong \mathbf{Z}_5 \times \mathbf{Z}_{11} \times \mathbf{Z}_{61}$, which is distinct from all other cases since for these cases $G$ is nonabelian.

To see that the second and third cases are distinct, let $\langle x, y, z \rangle$ denote the second case and let $\langle X, Y, Z \rangle$ denote the third case. Assume the two are isomorphic, say $x \mapsto X^\prime$ and $y \mapsto Y^\prime$. Then $X^\prime$ is a generator of a $5$-Sylow subgroup, so that $X^\prime = GXG^{-1}$. By assumption, $$X^\prime Y^\prime \left( X^\prime \right)^{-1} = Y^\prime.$$ Since $Y^\prime$ is a generator of the unique $11$-Sylow subgroup, $Y^\prime = Y^\alpha \left( \alpha = 1, \dots, 10 \right)$. Also, by normality, $GYG^{-1} = Y^n \left( n = 1, \dots, 10 \right)$. Hence, \begin{align} Y^\alpha &= GXG^{-1}Y^{\alpha}GX^{-1}G^{-1} \\ &= GXY^{n\alpha} X^{-1}G^{-1} = G Y^{2n\alpha} G^{-1} = Y^{2n^2\alpha}. \end{align} Hence $1 = Y^{(2n^2 - 1)\alpha}$. So we have $11 \mid (2n^2 - 1)\alpha$ implying $11 \mid 2n^2 - 1$. One can check that $11 \nmid 2n^2 -1$ for $n = 1, \dots, 10$. Thus we arrive at a contradiction.

It is interesting that there are solutions to problems of the form \begin{align} p \mid 2n^2 - 1 && \left( p \text{ prime}, n = 1, \dots, p - 1 \right). \end{align} For instance $p = 7$ and $n = 2$ is a solution. This means that the method above need not have yielded a contradiction as presented.

To me, this illuminates some subtlety in questions of this type, for we see an instance where isomorphism types depend on some number-theoretic calculations. It begs the question whether it is reasonable to be so certain (based on the discussion so far) that there aren't fewer than seven isomorphism types for this problem.

If anyone sees a more obvious reason why, e.g., the second and third cases are non-isomorphic, I would be interested to hear it. Later I might get to examining some of the other cases.