Let $G$ be a group of order $16$, and $\pi$ a $G$ action on a set $X$, where $|X|=5$.
We also know that in $G$ there are only $5$ sub-groups of order $4$.
Prove that $G$ has a non-trivial normal sub-group.
Any hints/ideas?
Asked
Active
Viewed 160 times
0
1 Answers
1
This is a $p$-group as it's order is of the form $p^k$(precisely $2^4$), and center of a $p$-group is non-trivial and center is a normal subgroup, hence done!
Arpan1729
- 3,414
-
Does it matter that there are only $5$ sub-groups of order $4$? – ChikChak May 21 '17 at 09:32
-
No, not required – Arpan1729 May 21 '17 at 09:38
-
I suspect that this answer would/will not get full marks. There is a strong implication that you are supposed to prove it by using the fact that there are (exactly) $5$ subgroups of order $4$, which is possible. – Derek Holt May 21 '17 at 12:26
-
Sorry but why does G have $5$ subgroups of order $4$? – Nour May 26 '17 at 19:20