In [Huppert, Endliche Gruppen, p140] the author shows that the alternating group $A_5$ is isomorphic to $G := \langle x,y \mid x^5=y^2=(xy)^3=1 \rangle$. The proof is elementary but long and complicated. Is there a simple way to prove the assertion by using some theory? Of course essentially we have to show that $|G| \leq 60$.

Here is a possible attempt: $A_5$ is generated by $(1,2,3,4,5)$ and $(12)(34)$, and these elements satisfy the above relations. We can try to give a proof of $|A_5| \leq 60$ by using these generators (and the well known subgroup structure of $A_5$), and then to adapt the same proof for $G$. This could be done as follows:

Set $a := xy$ and $b := (xy)^{x^2} = x^{-1}y{x^2}$. Both elements are of order three. The corresponding permutations are $(2,4,5)$ and $(1,2,4)$ so in principle we should be able to show that $U := \langle a,b \rangle$ (which is in fact isomorphic to $A_4$) has at most $12$ elements. For doing so we define $V := \langle ab, (ab)^b \rangle$. $V$ has to be isomorphic to the Klein four group, so we have to show that $(ab)$ and $(ab)^b$ are commuting involutions (should be possible somehow...), and that $b$ normalizes $V$ (easy). Then it is clear that $U = V \langle b \rangle$ has at most $12$ elements. Finally, we have to show that the index $|G:U|$ is at most $5$. This is the only part, where I have no idea how to proceed.

Any ideas?

  • 1
    $\begingroup$ The easiest way to show that $|G:U|=5$ is to apply the Todd-Coxeter coset enumeration algorithm. $\endgroup$
    – Derek Holt
    Nov 4, 2016 at 19:34
  • $\begingroup$ @DerekHolt Thanks! I was not aware of that algorithm, but luckily I was able to finish my proof without using it. $\endgroup$
    – Dune
    Nov 4, 2016 at 22:22
  • $\begingroup$ I prefer the use of the algorithm because it is equally easy, but also more systematic and of general applicability. Also, it is not difficult to prove the result using coset enumeration and then to convert the calculation into a proof like the one you gave. (I am talking here about the proof that $|G:U|=5$, not the proof that $|U| \le 12$. $\endgroup$
    – Derek Holt
    Nov 5, 2016 at 10:21
  • $\begingroup$ @DerekHolt: You're right, the Todd-Coxeter algorithm yields $|G:U|=5$ extremely fast in this case. Thank you for mentioning it! $\endgroup$
    – Dune
    Nov 6, 2016 at 18:45
  • $\begingroup$ Related : math.stackexchange.com/questions/1077664/… $\endgroup$
    – Arnaud D.
    Mar 2, 2018 at 13:49

1 Answer 1


Finally, I am able to complete my sketch of the proof. We begin by proving the following:

$G := \langle x,y \mid x^3=y^3=(xy)^2 = 1 \rangle$ is isomorphic to $A_4$

Proof: $A_4$ is generated by $(123)$ and $(234)$, and these permutations satisfy the above relations. Hence, $A_4$ is a homomorphic image of $G$. We will show henceforth $|G| \leq 12$. Let $a = xy$ and $b = a^x = yx$. We have $a^2 = b^2 = 1$, and also $(ab)^2 = xy^{-1}x^{-1}y^{-1}x = x (xy)^{-2}x^2 = 1$. So $V := \langle a,b \rangle$ is a homomorphic image of $C_2 \times C_2$. Since $a^x = b \in V$ and $b^x = x^{-1}yx^2 = (yx)^{-1}(xy)^{-1} = ba \in V$, $\langle x \rangle$ normalizes $V$, and $G = V \langle x \rangle$ has at most 12 elements. $\square$

Now we are able to prove the original statement:

$G := \langle x,y \mid x^5=y^2=(xy)^3=1 \rangle$ is isomorphic to $A_5$

Proof: $A_5$ is generated by $(12345)$ and $(12)(34)$, and these permutations satisfy the above relations. Hence, $A_5$ is a homomorphic image of $G$. We will show $|G| \leq 60$. Let $a = xy$ and $b = a^{x^2} = x^{-1}yx^2$. We have $a^3=b^3=1$. In the following we will frequently need the identity

$$yx^{-1}y= xyx,\tag{$\ast$}$$

which follows directly from $(xy)^3=1$. Using $(*)$ we compute $(ab)^2 = x(yx^{-1}y)x^3(yx^{-1}y)x^2 = x(xyx)x^3(xyx)x^2 = 1$. Hence, $U := \langle a,b \rangle$ is a homomorphic image of $A_4$, and has therefore at most $12$ elements. We finish the proof by showing that the complete set of right cosets of $U$ in $G$ is given by $\Omega = \{ U, Ux, Ux^2, Ux^3, Ux^4 \}$. Since $G$ acts transitively on its right cosets, this can be done by showing that $\Omega$ is invariant under the action of the generators $x$ and $y$. It is clear that $\Omega x = \Omega$. Furthermore, we have

  • $Uy = Uay = Ux$
  • $(Ux)y = Ua = U$
  • $(Ux^2)y = Ux(xyx)x^{-1} = Ux(yx^{-1}y)x^{-1} = Uabx^2 = Ux^2$
  • $(Ux^3)y = Ub^{-1}x^4 = Ux^4$
  • $(Ux^4)y = Ubx^3 = Ux^3$

This also shows $\Omega y = \Omega$, and hence $\Omega G = \Omega$, which completes the proof. $\square$

I am quite satisfied with this proof, since it is very conceptual. But still it is quite long and depends on many calculations which seem a bit random. I would be happy to see shorter proofs which are using more sophisticated concepts.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .