if a is the only element of order k in G, then a is in the center of G

Question

This is a question from A Book Of Abstract Algebra. I found this question which gives the same hint the book gives. Here are the hints given by the book:

$ord(a) = ord(bab^{-1})$
if $a = bab^{-1}$, then $ab = ba$ (i.e. $a$ and $b$ commute).

I still don't understand how to solve this. $ord(a) = ord(bab^{-1})$ doesn't mean $a = bab^{-1}$, right?

You are right in general, but if $\text{ord}(a) = \text{ord}(bab^{-1})$ and $a$ is the only element of its order... — Kaj Hansen, Dec 15 '16 at 00:52

score 1 · Accepted Answer · answered Dec 15 '16 at 00:52

1

If $\operatorname{ord}(a) = \operatorname{ord}(bab^{-1})$, then since $a$ is the only element which has the same order as $a$ (unique by its order), then $bab^{-1}$ must equal $a$.

answered Dec 15 '16 at 00:52

Tom

9,978

score 0 · Answer 2 · answered Dec 15 '16 at 00:52

0

Hint If $ord(a)=k$ then $ord(bab^{-1})=k$. Since $a$ is the only element of order $k$ in the group, it means that any element of order $k$ must be equal to $a$.

answered Dec 15 '16 at 00:52

N. S.

132,525

if a is the only element of order k in G, then a is in the center of G

2 Answers2