1

I have a question on the proof of: Proving if $|G|=280$, then $G$ is not simple

$n_7=8$ so there are $48$ elements of order $7$.

$n_5=56$ so sthere are $224$ elements of order $5$.

Now, in the proof, the $8$ elements left belong to the $2$ Sylow group, so it follows that there is only one $2$ Sylow subgroup, which is normal, too. How does it follow?

Is it also right, if I argue with there are at most $35$ elements of order $2$, so there are $48+224+35+1=308>280$ elements, so $G$ can't be simple?

Shaun
  • 47,747
Gerturter
  • 195

1 Answers1

2

Every element which lies in some Sylow $2$-subgroup has order equal to a power of $2$. Since the Sylow $2$-subgroups of $G$ have $8$ elements each, if there was more than one, then there would be more than $8$ elements of order a power of $2$, contradicting what was already established.

Wojowu
  • 27,526
  • But is it also okay to show that there are at most $35$ elements of order $2$ so that $G$ can't be simple? – Gerturter Jan 25 '20 at 11:37
  • @Gerturter I'm not sure how you argue that - how do you know there are at most $35$ such elements, and how do you derive a contradiction from that? (in your question it sounds like you are using there are at least 35 such elements) – Wojowu Jan 25 '20 at 13:02
  • If I consider $n_2$ then it's $n_2 \in \lbrace 1, 5, 7, 35\rbrace$, so there are at most $35$ elements of order $2$. Then $|G|<48+224+1+35$, so it's not simple. – Gerturter Jan 25 '20 at 13:37