# Centre of a group proof

• May 4th 2012, 08:16 PM
Jesssa
Centre of a group proof
Hey,

I have this question which asks to prove that if a group has an element, a such that |a|=2, (exactly one order 2 element a) that this element is in the centre of the group.

so ag=ga for all g in G

aga-1=g, aa-1g=g so aa-1=e which is true since a has order 2.

But I feel like this isn't enough at all and I cant really see where the "exactly 1 element with |a|=2" comes in,

Does anyone have any ideas?
• May 5th 2012, 12:03 AM
Sylvia104
Re: Centre of a group proof
You're supposed to prove that $ag=ga$ for all $g\in G,$ not assume $ag=ga$ and play around with it (a common mistake with many students). (Shake)

This is what you do. Let $b=g^{-1}ag.$ Show that $b$ order $2.$ But $a$ is the unique element of order $2$ in $G.$ What does this tell you? (Smile)
• May 5th 2012, 12:16 AM
Jesssa
Re: Centre of a group proof
So you can say that since a is in the group then its conjugate must also be in the group, let b be its conjugate then

bb=g-1agg-1ag= g-1aag=g-1g= e so b also had order 2

Since a is the only element of order to b must be a

Does that look alright?
• May 5th 2012, 12:55 AM
Sylvia104
Re: Centre of a group proof
That's correct. One more line completes the proof. (Nod)
• May 5th 2012, 01:13 AM
Jesssa
Re: Centre of a group proof
Thanks Sylvia!

Do you mean like, for the final line, a conclusion like,

Therefore as b = a, since the group has only one order 2 element,

b = g-1ag
gb=ag
ga=ag

and hence a is in the centre of G