# Thread: Order of Group Elements

1. ## Order of Group Elements

Hi, I need help solving these problems:

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

1. If ak =e where k is odd, then the order of a is odd.

3.The order of ab is the same as the order of ba. (Hint: if (ba)n = (baba…..b)a = e. Let (baba….b) =x then a is the inverse of x. Thus, ax =e.)

2. Let x =a1a2… an, and let y be a product of the same factors, permuted cyclically. (That is, y = akak+1…ana1…ak-1.) Then ord(x) = ord(y).

Thanks!

2. Originally Posted by steph615
1. If ak =e where k is odd, then the order of a is odd.
You probably mean $a^k = e$. If $d$ is the order of $a$ then $d|k$. So $d$ cannot be even, i.e. it must be odd.

3.The order of ab is the same as the order of ba. (Hint: if (ba)n = (baba…..b)a = e. Let (baba….b) =x then a is the inverse of x. Thus, ax =e.)
$(ab)^n = 1 \implies (ab)...(ab) = 1 \implies a (ba)^{n-1} b = 1 \implies (ba)^{n-1} = a^{-1}b^{-1}$
But $a^{-1}b^{-1} = (ba)^{-1}$ thus $(ba)^{n-1} \implies (ab)^{-1} \implies (ba)^n = 1$.
(Why is this not a complete proof?)

2. Let x =a1a2… an, and let y be a product of the same factors, permuted cyclically. (That is, y = akak+1…ana1…ak-1.) Then ord(x) = ord(y).
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Use induction, with the basic case covered in #2.

3. ## Re

Ok, the proof for ord(ab)=ord(ba) is not complete because don't you have to show that n is the smallest positive integer? How then would I show that?