# finite group of odd order

• October 15th 2010, 04:38 AM
vincent
finite group of odd order
let G be a group has a finite number of elemenys, half of them have order 2 and the other half order different than 2, show that the subgroup H of the elements of order other than 2 has an odd number of elements and that it is a comutative subgroup.

so far i know how to prove that a group has an odd number of elements iif it has no element of order 2 and since H has index 2 in G, it must be normal in G because its left cosets will be equal to its right cosets since it has only 2, itself and the other one.

any help will be greatly apreciated.
• October 15th 2010, 06:39 AM
Swlabr
Quote:

Originally Posted by vincent
let G be a group has a finite number of elemenys, half of them have order 2 and the other half order different than 2, show that the subgroup H of the elements of order other than 2 has an odd number of elements and that it is a comutative subgroup.

so far i know how to prove that a group has an odd number of elements iif it has no element of order 2 and since H has index 2 in G, it must be normal in G because its left cosets will be equal to its right cosets since it has only 2, itself and the other one.

any help will be greatly apreciated.

Can you come up with an example? I suspect such a group must be the group $C_2 = \mathbb{Z}/2\mathbb{Z}$, the cyclic group of order 2. If our group has order 2n, clearly the elements of odd order (aka those not of order 2) form a group, but so the the elements of order 2 unioned with the identity. These groups have order n and n+1 respectively, and so n+1 divides 2n...

EDIT: That is, if you can prove that the product of two elements of order 2 in a finite group must have even order, but this is true by Sylow's Theorem...which you may or may not have done yet...
• October 15th 2010, 10:27 AM
Bruno J.
Hi there vincent! :)

Swlabr : consider $S_3$. Each of $(12), (13), (23)$ has order $2$; the other three elements $1, (123), (132)$ form a subgroup isomorphic to the cyclic group of three elements. So that's another example.

It is not true that the product of two elements of even order has even order. For example $(12)(23)=(123)$ has order $3$.
Also you say
Quote:

the elements of odd order (aka those not of order 2)
but there could be an element of order not $2$ which is not of odd order!

Vincent : are you given that the elements of order other than $2$ form a subgroup? Because in general that's not true. For example in $Z_4=$, $x$ has order $4$ (which is not $2$) but $x^2$ has order 2.
• October 15th 2010, 12:47 PM
Swlabr
Quote:

Originally Posted by Bruno J.
Hi there vincent! :)

Swlabr : consider $S_3$. Each of $(12), (13), (23)$ has order $2$; the other three elements $1, (123), (132)$ form a subgroup isomorphic to the cyclic group of three elements. So that's another example.

It is not true that the product of two elements of even order has even order. For example $(12)(23)=(123)$ has order $3$.
Also you say but there could be an element of order not $2$ which is not of odd order!

Vincent : are you given that the elements of order other than $2$ form a subgroup? Because in general that's not true. For example in $Z_4=$, $x$ has order $4$ (which is not $2$) but $x^2$ has order 2.

Fair point about the two elements of order two not multiplying together to give you an element of order two, I hadn't thought through my Sylow's Theorem thought (it just means they're conjugate).

However, in the circumstances we are in every element of even order has order 2, as if the elements not of order 2 form a subgroup and there is an element, a, of order 2n in it then $a^n$ has order 2...
• October 15th 2010, 08:25 PM
vincent
Bruno: yes we are given that the elements of order not 2 forms a subgroup, I had forgotten about that part...
• October 16th 2010, 01:02 PM
Bruno J.
Here's my solution!

Spoiler:
Let $H$ be the subgroup of elements of order not $2$. Pick $a \in G-H$, so $a=a^{-1}$. We can write $G=H \coprod aH$ (disjoint union). Let $f: H\to H$ be the homomorphism $x \mapsto axa^{-1}=axa$. (The range really is $H$ because $H \triangleleft G$.) We have $f(x)x=axax=(ax)^2=1$ for every $x \in H$ (because $ax \in G-H$). Hence $f(x)=x^{-1}$, and $f(xy)=(xy)^{-1}=f(x)f(y)=x^{-1}y^{-1}$, so $xyx^{-1}y^{-1}=1$, i.e. $x$ and $y$ commute for any $x,y \in H$.