True or false: the group of invertible $2\times 2$ diagonal matrices over $\mathbb{Z}_7$ is cyclic. I just want to make sure my reasoning is correct.

Since there are $36$ such matrices ($6$ choices for $a_{1,1}$ and $6$ for $a_{2,2}$). We know this groups is abelian and thus has a subgroup for every $k\in \mathbb{Z}^+$ where $k$ divides $36$. There are two groups (up to isomorphism) with order $6$, namely $\mathbb{Z}_6$ and $S_3$. Now $\mathbb{Z}_6$ has zero divisors ex: $2\cdot 3=6=0$, but our matrices are invertible and thus we have no zero divisor. Also, $S_3$ is not cyclic. Thus our group is not cyclic.


1 Answer 1


I'm afraid that the reference to $S_3$ and zero divisors is not valid.

There are indeed two subgroups having order $6$, but both are cyclic (and isomorphic to each other): one consists of the matrices of the form $$ \begin{bmatrix} x & 0 \\ 0 & 1 \end{bmatrix} $$ and the other one with the elements along the diagonal switched.

In a cyclic group there is exactly one subgroup for each divisor of the order.

More easily, for every invertible matrix of your group, we have $$ \begin{bmatrix} x & 0 \\ 0 & y\end{bmatrix}^6= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} $$ so no element can generate a group with $36$ elements.

  • $\begingroup$ I never said S3 does not have zero divisors, I said it was not cyclic; thus its parent groups cant be cyclic. $\endgroup$
    – tmpys
    Commented Sep 17, 2017 at 21:21
  • 2
    $\begingroup$ @tmpys There's no way to embed $S_3$ in the given group, which is abelian. The subgroup of matrices of the form I mentioned is cyclic of order $6$, but with respect to matrix multiplication, and you're using multiplication on $\mathbb{Z}_6$ under which it is not a group. You're comparing apple and oranges. $\endgroup$
    – egreg
    Commented Sep 17, 2017 at 21:28
  • $\begingroup$ ahhhhhh! I see. Thanks. $\endgroup$
    – tmpys
    Commented Sep 17, 2017 at 21:47

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