Prove that a group that has more than one subgroup of order 5 must have order at least 25.

Question

I know that my group would have to have an order divisible by 5 but I don't know the best way of solving this question. I first assumed I had two subgroups $$|H|\, and\,|K|\, with\, order\, 5\, but\, H\, \neq K $$ I am not sure where to go after this.

Any help would be appreciated.

score 2 · Accepted Answer · answered Apr 11 '20 at 19:29

2

Hint: for the set $HK$ we have $|HK|=\frac{|H||K|}{|H \cap K|}$ and by Lagrange $H \cap K=1$.

answered Apr 11 '20 at 19:29

Nicky Hekster

49,281

Why would $$H \cap K = 1$$ – Dmot Apr 11 '20 at 19:35
$H \cap K$ is a subgroup of $H$ (and $K$). Hence its order divides $5$. If it is $5$ then $H \cap K=H$, which is equivalent to $H=K$. – Nicky Hekster Apr 11 '20 at 19:39

Prove that a group that has more than one subgroup of order 5 must have order at least 25.

1 Answers1