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Let a,b, and c be elements of a group G. Prove the following:

If $a^k$=e where k is odd, then the order of $a$ is odd

Here is how I worked on the proof:

Assume $a^k$=e where k is odd. Since k is odd then k=2L+1 for L $\in \mathbb{Z}$.

Therefore, through replacement:

$a^k$=$a^{2L+1}$=$a^{2L} \times a$=e

I'm really tempted to say since L is any integer, $a^{2L}$ will always be even while a is odd and the result will be odd but i'm thinking in terms of permutations. The problem I'm having is I'm not sure how to connect it, well...that is if I am on the right track in the first place.

mika
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2 Answers2

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Let $o(a)=m$

then $m$ divides $k$ ; since $k$ is odd $m$ must be odd

Learnmore
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  • why down vote for this answer? – Learnmore Mar 14 '15 at 08:10
  • I did not downvote. I just worked on this exercise on my own. Does this reasoning make sense:

    Since $n$ must be non-zero, the first legal $n$ value is $1$. Because $a^{k}=e$ when $k$ is odd, and $1$ is odd, then $a$'s order must be $1$, i.e. odd.

     – Kevin Meredith Jun 06 '15 at 22:04
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If the order of $a$ is $m$ and $a^k=e$, then consider the quotient $q$ and remainder $r$ when dividing $k$ by $m$ so $k=mq+r$ with $0 \le r \lt m$.

You then have $e=a^{mq+r}=a^{mq} \,a^r=(a^m)^q \,a^r = e^q\, a^r = a^r$, and since $r \lt m$ you must have $r=0$ as otherwise the order of $a$ would be $r$ rather than $m$.

If $r=0$ than $k$ is a multiple of $m$, so $k$ being odd implies $m$ must be odd since all multiples of even numbers are even.

Henry
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