1
$\begingroup$

If the following question is not good enough, please close it.

Please give a elementary proof or reference of the following result:

Let $G$ be a finite groups of order $120$, and $|Z(G)|=2$. Suppose $G/Z(G) \cong A_5$ and $G'=G$, then $G \cong SL(2,5)$.

Of course, using the Schur index of $A_5$, I can get the proof. But how to prove a group to be a classical group is always trouble for me. I know this result of may be well-known for a long time ago, but I can not find the proof in any textbook. (I study this because that I want to know more about groups of order $8pq$.

$\endgroup$
1
  • 1
    $\begingroup$ The conditions that ${\rm Z}(G)$ has order 2 and that $G/{\rm Z}(G) \cong {\rm A}_5$ are redundant and can be left away. Groups of such small size can easily be looked up in a standard database (e.g. GAP's Small Groups Library), which you might cite if you need a reference. Among others, GAP's Small Groups Library contains all groups of order $\leq 2000$ except for those of order 1024. There you can simply test all groups of a particular order whether they satisfy your conditions. $\endgroup$
    – Stefan Kohl
    May 13, 2013 at 15:01

2 Answers 2

4
$\begingroup$

Since you know how to prove this already, it is difficult to know what you are looking for!

Here is a quick proof that there is a unique isomorphism class of groups $G$ with this property using the well-known fact that $A_5$ has the presentation $\langle x,y \mid x^2=y^3=(xy)^5=1 \rangle$.

Let $t$ generate the centre of your group $G$, and let $X,Y$ be inverse images in $G$ of generators $x,y$ of $A_5$. By replacing $Y$ by $Yt$ if necessary, we can assume that $Y^3=1$. By replacing $X$ by $Xt$ if necessary we can assume that $(XY)^5=1$. If $X^2=1$, then $X$ and $Y$ generate a subgroup isomorphic to $A_5$, which must be a complement of $Z(G)$, contradicting $G'=G$. So $X^2=t$ and hence

$$G \cong \langle X,Y,t \mid t^2=1, X^2=t, y^3=(XY)^5=1, tY=Yt \rangle.$$

$\endgroup$
0
3
$\begingroup$

Another way to prove this is using ordinary character theory and Brauer character theory, though it needs a little more background. I outline a proof, since it illustrates many more general facts in a nice way. Since $G = G^{\prime},$ every complex irreducible character of $G$ which does not contain $Z(G)$ in its kernel has even degree. If the degrees of the irreducible characters not containing $Z(G)$ in their kernels are $2m_{1},\ldots,2m_{t},$ possibly with repetition, we have $\sum_{i=1}^{t}m_{i}^{2} = 15.$ Hence at least $3$ of the $m_{i}$ are odd, and we may suppose that $m_{1} = 1,$ as claimed. Now $G$ contains a unique involution, as its two dimensional complex irreducible character is faithful. Hence $G$ has a quaternion Sylow $2$-subgroup of order $8.$ Any element of order $6$ in $G$ is conjugate to its inverse. Hence the value of the irreducible character of degree $2$ on $5$-regular elements is rational. But an irreducible Brauer character is realizable over the field of its character, and the $2$-dimensional irreducible complex representation of $G$ clearly remains absolutely irreducible on reduction (mod $5$). Hence $G$ is isomorphic to to a subgroup of ${\rm GL}(2,5).$ Since $G = G^{\prime},$ we see that $G$ is isomorphci to a subgroup of ${\rm GL}(2,5)^{\prime} = {\rm SL}(2,5).$ Since $|G| = 120,$ we have $G \cong {\rm SL}(2,5).$

A similar argument using the $4$-dimensional complex representation of a double cover of $A_{7}$ shows that $A_{7}$ is isomorphic to a subgroup of ${\rm GL}(4,2)$, and the index is easily seen to be $8.$ This gives an embedding of ${\rm GL}(4,2)$ into $A_{8},$ which is an isomorphism on consideration of order, thus exhibiting the well-known "exceptional" isomorphism $A_{8} \cong {\rm GL}(4,2).$

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.