# Thread: Prove that if b is the only element of order 2 in G, then b E Z(G)

1. ## Prove that if b is the only element of order 2 in G, then b E Z(G)

the center Z(G) is defined as Z(G) = {a E G | ax = xa for every x E G}
Prove that if b is the only element of order 2 in G, then b E Z(G)

2. Originally Posted by rainyice
the center Z(G) is defined as Z(G) = {a E G | ax = xa for every x E G}
Prove that if b is the only element of order 2 in G, then b E Z(G)
This is actually quite a neat question. What you need to do it to prove that $\displaystyle b$ satisfies this condition. I will just give you a hint though. Look in this thread of yours. You will use one of these three facts, along with the fact that if $\displaystyle x^2=e$ then $\displaystyle x=b$. You will also have to fiddle around with the condition. $\displaystyle ax=xa \Leftrightarrow \ldots$

3. Then G has only one subgroup of order 2, then $\displaystyle <b>$ is normal in G since it doesn't have any conjugates. So then if $\displaystyle x\neq e$, then $\displaystyle xbx^{-1}=b$ since it must be in $\displaystyle <b>$ but it's not $\displaystyle e$. QED