I'm asked to show that $G=GL_2(\Bbb Z/2\Bbb Z)$ is isomorphic to $S_3$. I have few ideas but I don't manage to put them all together in order to obtain a satisying answer. I first tried using Cayley's theorem ($G$ is isomorphic to a subgroup of $S_6$), and I also noticed that $\operatorname{Card}(G)=\operatorname{Card}(S3)=6$ & that they're both non-abelian group.

Is this enough to say that considering $S_3$ is a subgroup of $S_6$ with the same cardinality than $G$, it has to be isomorphic to it ? Could anyone give me some elements to get a more rigorous proof or lead me to an other path to show this statement ? Thanks in advance

  • $\begingroup$ It is not enough to say that. For instance, $\mathbb{Z}_4$ has $4$ elements and so does $\mathbb{Z}_2 \oplus \mathbb{Z}_2$, but they are not isomorphic. Even if things have the same order and are contained within the same group, this is not sufficient. $\endgroup$ Nov 14, 2013 at 17:41
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    $\begingroup$ @mathematics2x2life Except in this case there is only one non-abelian group of the appropriate order, so it's perfectly reasonable to say '$G$ is of order 6 and is non-abelian so it's $S_3$' - assuming that both of those statements HAVE been proven... $\endgroup$ Nov 14, 2013 at 17:48
  • $\begingroup$ Not all 6-element subgroups of $S_6$ are isomorphic. Some are isomorphic to $S_3$, but you could also have a 6-cycle, so your approach doesn't quite work. $\endgroup$ Nov 14, 2013 at 17:51
  • $\begingroup$ @StevenStadnicki Which I am in total agreement with. However, this comes from a stronger fact ( a fortiori) and you'd need to either reference that fact or prove it. It wouldn't be enough to say, 'they have the same order, QED'. In general, this would not be the case. Here, you'd need the fact that they have the same order AND the fact that there can only be $1$ nonabelian group of order $6$. $\endgroup$ Nov 14, 2013 at 17:51
  • $\begingroup$ Simplify with $\mathbb{Z}_2 \cong \mathbb{Z/2Z}$ $\endgroup$ Jan 10, 2017 at 19:58

2 Answers 2


If you don't know that the unique non-abelian group of order $6$ is $S_3$, you can go with a more explicit approach:

Find an isomorphism by showing that $GL_2(\mathbb Z/2\mathbb Z)$ is a group of permutations of a $3$-element set, the nonzero vectors in $\mathbb (\mathbb Z/2\mathbb Z)^2$. Number these vectors, then argue that every element of $GL_2$ induces a permutation of this set, which gives you a homomorphism between the two groups by sending each element of $GL_2$ to the corresponding permutation in $S_3$. Then show that this homomorphism is injective: two different elements of $GL_2$ permute the elements differently. Finally, count the number of elements in each group. Since the groups have the same order, your injective homomorphism is also surjective, and thus an isomorphism.

  • $\begingroup$ Thanks a lot for your answer. I'm going to try to prove it with your method too. $\endgroup$ Nov 14, 2013 at 17:56
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    $\begingroup$ Great. I left quite a few gaps for you to fill in, so I'm glad you're going to give it a shot. Post a comment if you get stuck. $\endgroup$ Nov 14, 2013 at 17:58
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    $\begingroup$ @BrettFrankel: +1 but what about showing that they are the same presentations? $\endgroup$
    – Mikasa
    Nov 14, 2013 at 18:02
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    $\begingroup$ @B.S. Working with generators and relations would be another solution altogether, which you could certainly write up if you are so inclined. $\endgroup$ Nov 14, 2013 at 18:09

You can do this in 2 steps :

  1. $GL_2(\mathbb{Z}/2\mathbb{Z})$ has order 6, and is non-abelian.
  2. Any non-abelian group of order 6 is isomorphic to $S_3$

To prove 2 : Let $G$ be any non-abelian group of order 6, then by Cauchy's theorem, there is a subgroup $H<G$ of order 2 and a subgroup $K$ of order 3. Since $[G:K] = 2$, $K$ is normal in $G$. If $H$ were normal in $G$, then

(a) $G = HK$

(b) $H\cap K = \{e\}$

(c) For all $h\in H, k\in K, hkh^{-1}k^{-1} \in H\cap K = \{e\}$

From (a), (b) and (c), you could conclude that $G$ is abelian, which it is not.

Hence, $H$ is not normal in $G$. Now let $G$ act on the set of left cosets of $H$ in $G$. This would give a homomorphism $$ f: G\to S_3 $$ Now check that $\ker(f) < H$, and since $H$ is not normal in $G$, $\ker(f) \neq H$. The only possibility is that $\ker(f) = \{e\}$. Hence $f$ is injective. Since $$ |G| = |S_3| = 6 $$ it follows that $f$ is surjective as well, and hence $f$ gives an isomorphism $G\cong S_3$

Have you seen either of these facts? If not, then say so in the comments and I can indicate the proofs.

  • $\begingroup$ Thanks for your answer. I know how to "prove" your first point, but I'd be very grateful if you could help me with the second one. $\endgroup$ Nov 14, 2013 at 17:51
  • $\begingroup$ @MartinPicot : I have added a proof of (2) $\endgroup$ Nov 14, 2013 at 17:56

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