T o prove that if $n=2k$ and $1\leq i < n $ then $x^{2i}=e$ iff $i=k$

Question

Let $x$ be element of finite order $n$ in G.

To prove that if $n=2k$ and $1\leq i < n $ then $x^{2i}=e$ iff $i=k$

Now converse part i trivial. Now to prove that if $n=2k$ and assuming $x^{2i}=e$ i have to prove that $i=k$, Assuming contradiction that $i \neq k$ and so $i<k$ and thus $2i<2k$. This contradicts fact thatx has order $2k$.

Is this correct?

Thanks

It's correct, albeit it's not really a proof by contradiction. Sort of contrapositive rather (argument of type ‘otherwise…’). — Bernard, Aug 06 '16 at 13:08
that's correct but to be more pedanic, you should show the i>k case. — Zau, Aug 06 '16 at 13:10

score 0 · Answer 1 · edited Apr 13 '17 at 12:20

0

Hint $\ $ If $\,x\,$ has order $\,2k\,$ then $\,x^{2i}=1\iff 2k\mid 2i\iff k\mid i$

edited Apr 13 '17 at 12:20

Community

1

answered Aug 06 '16 at 13:10

Bill Dubuque

272,048

1

is the op's proof correct? – Zau Aug 06 '16 at 13:11
@Zack Has already been adfressed in comments, but can also be inferred by the Hint (esp. if you follow the link, which leads to a conceptual explanation of the essence of the matter). – Bill Dubuque Aug 06 '16 at 13:25

T o prove that if $n=2k$ and $1\leq i < n $ then $x^{2i}=e$ iff $i=k$

1 Answers1