abhishekkgp
New member
- Joined
- Jan 23, 2012
- Messages
- 25
The following question has two parts to it. I could solve part (a). Need help with part (b).
QUESTION: Let \(\displaystyle G\) be a group of order \(\displaystyle 30\).
a) Show that a 3-sylow subgroup or a 5-sylow subgroup is normal in \(\displaystyle G\).
b) From part (a) show that every 3-sylow subgroup and every 5-sylow subgroup is normal in \(\displaystyle G\).
ATTEMPT:
(a) \(\displaystyle 30=2 \times 3 \times 5\).
The possible values of \(\displaystyle n_5\) are \(\displaystyle 1\) and \(\displaystyle 6\).
The possible values of \(\displaystyle n_3\) are \(\displaystyle 1\) and \(\displaystyle 10\).
Assume no 3-sylow subgroup and no 5-sylow subgroup is normal in \(\displaystyle G\). This means \(\displaystyle n_5=6, n_3=10\).
This gives that there are at least \(\displaystyle (5-1)\times 6 + (3-1) \times 10 + 1= 45\) distinct elements in \(\displaystyle G\). This is clearly a contradiction. Therefore at least one of \(\displaystyle n_3\) and \(\displaystyle n_5\) is \(\displaystyle 1\). Thus this part of the question is solved.
What do i do to solve part (b).
QUESTION: Let \(\displaystyle G\) be a group of order \(\displaystyle 30\).
a) Show that a 3-sylow subgroup or a 5-sylow subgroup is normal in \(\displaystyle G\).
b) From part (a) show that every 3-sylow subgroup and every 5-sylow subgroup is normal in \(\displaystyle G\).
ATTEMPT:
(a) \(\displaystyle 30=2 \times 3 \times 5\).
The possible values of \(\displaystyle n_5\) are \(\displaystyle 1\) and \(\displaystyle 6\).
The possible values of \(\displaystyle n_3\) are \(\displaystyle 1\) and \(\displaystyle 10\).
Assume no 3-sylow subgroup and no 5-sylow subgroup is normal in \(\displaystyle G\). This means \(\displaystyle n_5=6, n_3=10\).
This gives that there are at least \(\displaystyle (5-1)\times 6 + (3-1) \times 10 + 1= 45\) distinct elements in \(\displaystyle G\). This is clearly a contradiction. Therefore at least one of \(\displaystyle n_3\) and \(\displaystyle n_5\) is \(\displaystyle 1\). Thus this part of the question is solved.
What do i do to solve part (b).