Re: Centre of a group proof

You're supposed to *prove* that $\displaystyle ag=ga$ for all $\displaystyle g\in G,$ not assume $\displaystyle ag=ga$ and play around with it (a common mistake with many students). (Shake)

This is what you do. Let $\displaystyle b=g^{-1}ag.$ Show that $\displaystyle b$ order $\displaystyle 2.$ But $\displaystyle a$ is the unique element of order $\displaystyle 2$ in $\displaystyle G.$ What does this tell you? (Smile)

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^{-1}agg^{-1}ag= g^{-1}aag=g^{-1}g= e so b also had order 2

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

Does that look alright?

Re: Centre of a group proof

That's correct. One more line completes the proof. (Nod)

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