0

Prove that any simple group $G$ of order $60$ is isomorphic to $A_5$.

I am studying abstract algebra from Artin and studying these assignment from a masters level math course.

It gives hint.

Use the result :

if $H < G$, then $[ G : H]\geq 5$; if $[G : H] = 5$ then $G$ is isomorphic to $A_5$. The assumption that $G$ has no subgroup of index $5$ leads to contradiction.

I am sorry to say but I am not able to follow the hint. Why should $[G:H] \geq 5$ and why $[G:H] = 5$ then $G$ must be isomorphic to $A_5$? and why assumption $G$ has no subgroup of index $5$ leads to contradiction?

Can you please shed some light on this? Any other method is also welcome.

Shaun
  • 47,747
  • @lulu It's a typo . It should be greater than and I corrected it. I am sorry –  Sep 02 '21 at 13:16
  • If you have a subgroup $H$ with $[G:H] = 5$, then $G$ acts on the $5$-element set $G/H$ and this action gives you a morphism $G \to S_5$ which is injective since $G$ is simple. Since $|S_5| = 120$ the image of this morphism is a subgroup of $S_5$ of index $2$ and the only such subgroup is $A_5$. – Matthias Klupsch Sep 02 '21 at 13:36
  • Clearly, $G$ cannot have a subgroup of index $2$, because any subgroup of index $2$ is normal. It remains to show that $G$ cannot have a subgroup of index $3$ or $4$ either. – Geoffrey Trang Sep 02 '21 at 14:01
  • 1
    A likely duplicate of this. I won't vote that way right away, because there is some margin for disagreement (and my vote would be immediately binding). Anyway, aspects of this proof are covered there and in other threads linked to it. Do check them out! – Jyrki Lahtonen Sep 02 '21 at 15:30
  • See also this duplicate. – Dietrich Burde Sep 02 '21 at 16:53

0 Answers0