# Re: Finite group of order 4n+2 then elements of odd order form a subgroup.

• December 5th 2012, 03:07 AM
abhishekkgp
Re: Finite group of order 4n+2 then elements of odd order form a subgroup.
Let $G$ be a finite group of order $4n+2$ for some integer $n$. Let $g_1, g_2 \in G$ be such that $o(g_1)\equiv o(g_2) \equiv 1 \, (\mbox{mod} 2)$. Show that $o(g_1g_2)$ is also odd.
I found a solution to this recently but I think that solution uses a very indirect approach. Not saying that that solution was bad.. just indirect. So I wanted to see a more direct proof. I will post the solution I am talking about later in this thread once this is solved since if I post it now it might interfere with the thought process. I am sorry but I have no ideas of my own on how to go about doing it in a different way. Please help.
• December 5th 2012, 06:04 AM
ModusPonens
Re: Finite group of order 4n+2 then elements of odd order form a subgroup.
Since o(g_1) and o(g_2) are odd, they don't have a power of 2 in their prime decomposition. That means there are only odd numbers in their prime decomposition. Therefore , since o(g_1).o(g_2)=gcd(o(g_1), o(g_2)).lcm(o(g_1),o(g_2)), the number lcm(o(g_1),o(g_2)), which is o(g_1*g_2), has only odd prime factors, and thus is odd. I haven't used the hypothesis that the order of G is 4n+2, so does the question have subquestions?
• December 5th 2012, 07:06 AM
abhishekkgp
Re: Finite group of order 4n+2 then elements of odd order form a subgroup.
Quote:

Originally Posted by ModusPonens
Since o(g_1) and o(g_2) are odd, they don't have a power of 2 in their prime decomposition. That means there are only odd numbers in their prime decomposition. Therefore , since o(g_1).o(g_2)=gcd(o(g_1), o(g_2)).lcm(o(g_1),o(g_2)), the number lcm(o(g_1),o(g_2)), which is o(g_1*g_2), has only odd prime factors, and thus is odd. I haven't used the hypothesis that the order of G is 4n+2, so does the question have subquestions?

o(g_1g_2) can be entirely different than o(g_1)o(g_2). You probably have misread the question.
• December 5th 2012, 07:06 AM
Deveno
Re: Finite group of order 4n+2 then elements of odd order form a subgroup.
Quote:

Originally Posted by ModusPonens
Since o(g_1) and o(g_2) are odd, they don't have a power of 2 in their prime decomposition. That means there are only odd numbers in their prime decomposition. Therefore , since o(g_1).o(g_2)=gcd(o(g_1), o(g_2)).lcm(o(g_1),o(g_2)), the number lcm(o(g_1),o(g_2)), which is o(g_1*g_2), has only odd prime factors, and thus is odd. I haven't used the hypothesis that the order of G is 4n+2, so does the question have subquestions?

why is $|g_1g_2| = \text{lcm}(|g_1|,g_2|)$? i don't believe this is generally true (as far as i know, you can only assume this in ABELIAN groups where <g1> ∩ <g2> = {e}).

for example, in the dihedral group of order 10, with g1 = r, and g2 = r4, we have |g1g2| = |rr4| = |e| = 1, but lcm(5,5) = 5.
• December 6th 2012, 01:11 PM
ModusPonens
Re: Finite group of order 4n+2 then elements of odd order form a subgroup.
There's my mistake. Something was indeed fishy since i didn't use the hypothesis of the order of the group. Thanks Deveno and I apologise to abhishekkgp for giving a wrong answer.