# about the non-solvable group of order $120$

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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 $8pq$.

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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. –  Stefan Kohl May 13 '13 at 15:01

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

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

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Thank you very much, Prof. Holt. –  Wei Zhou May 13 '13 at 15:38

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 $A_{8} \cong {\rm GL}(4,2).$

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Thank you very much for the inspiring answer. –  Wei Zhou May 29 '13 at 12:58