Finding subgroups of $S_5$ with specific orders

Question

I want to verify whether there exist subgroups of $S_5$ of order $15,20,40$. but I don't know how to approach. generally, how do you prove when you are asked to find a subgroup with specific order of a finite group? any good/efficient way other than brute forcing with random generators?

Answer 1

$S_5$ has no subgroup of order $40$, see this duplicate for the arguments. The same holds for order $15$. For a subgroup of order $20$ we can take an element of order $5$, which exists by Cauchy, i.e., a $5$-cycle $(12345)$ and a $4$-cycle $(2354)$ to obtain a subgroup of order $20$. Here the $4$-cycle normalizes the subgroup generated by $(12345)$. In fact, the normalizer $N(P)$ of an order $5$ subgroup $P$ of $S_5$ has $20$ elements.

Answer 2

If it is just for verification and not for the exam/assignment, then you can also verify using computer. You can run the following GAP code:

G := SymmetricGroup( 5 );
List( AllSubgroups( G ), i -> Order( i ) );

After this you can visually verify, whether there is any subgroup of Order 15,20 and 40.

To get the actual subgroups (of order 20 in your case), you can use:

Filtered( l, i -> Order( i ) = 20 )

Hope this helps!

-- Mike