Subgroups of order $125$ in $S_{15}$

Question

I know symmetric group $S_{15}$ contains a copy of $C_5 \times C_5 \times C_5$ given by generators

$a=(1,2,3,4,5)$

$b=(6,7,8,9,10)$

$ c=(11,12,13,14,15)$

so $\langle a,b,c \rangle \cong C_5 \times C_5 \times C_5$. I am supposed to show every subgroup of $S_{15}$ of order $5^3=125$ is commutative and not normal in $S_{15}$. I have tried very hard in applying everything I know about symmetric groups and the Sylow theorems, but I am limited considering that $|S_{15}|=15!$. Help would be greatly appreciated!

Answer 1

The only normal subgroups of $S_n$ for $n\geq 5$ are: $\{e\}$, $A_n$, and $S_n$ itself.

The highest power of $5$ dividing $15!$ is: ?

Any two Sylow 5-subgroups of $S_{15}$ are conjugate in $S_{15}$ and, in particular isomorphic. The copy of $C_5\times C_5\times C_5$ in $S_{15}$ is a: ?

Conclusion: ?

I hope this helps!

Answer 2

Another way of looking at it is to determine the exponent of a group of order $125$. A simple argument on possible cycle decompositions gives that the exponent of such a group must be $5$.

Then consider the possible groups of order $125$. They are

• $C_{5}\times C_{5}\times C_{5}$;
• $C_{5}\times C_{25}$;
• $C_{125}$;
• $(C_{5}\times C_{5})\rtimes C_{5}$; and
• $C_{25}\rtimes C_{5}$.

The only one of these groups that has exponent $5$ is the first group.