# For any group $\mathrm{ord}( a_1\circ a_2\circ\dots\circ a_{n-1}\circ a_n)=\mathrm{ord}(a_2\circ a_3\cdots\circ a_{n-1}\circ a_n\circ a_1)$

Show through induction that in every Group G, for all $a_1,a_2,a_3,..., a_n$ that: $$\mathrm{ord}( a_1 \circ a_2 \circ ... \circ a_{n-1} \circ a_n) = \mathrm{ord}(a_2 \circ a_3 \cdots\circ a_{n-1} \circ a_n \circ a_1)$$

Me and some friends have talked about this. if the group was abelian, or if it was said that the operation is commutative, our job would have been done. however, we are pretty lost in this and basically are stuck on the "how do we proceed" department.

I appreciate any tips.

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Induction is unnecessary here: all you need to do is show that $\operatorname{ord}(ab)=\operatorname{ord}(ba)$, and then take $b=a_2a_3\ldots a_n$. – Chris Eagle May 9 '12 at 11:35
what do i do if i am being actually forced to use induction by the professor? – Shokodemon May 9 '12 at 11:36
I'm pretty sure that the proof Chris and lhf have both mentioned is the one expected, even though it isn't really induction. If you are really being forced to use induction, then you can just write the argument up in the form of an induction proof, though that is kind of silly. – Tara B May 9 '12 at 11:53
@Shokodemon: I added an answer showing how I'd write this as a proof by induction, but I think that lhf's answer is the one to accept, because it contains the actual proof. – Tara B May 9 '12 at 12:06

Let's prove something more general: $ord(xy) = ord(yx)$.

Suppose $ord(xy)=n$. Consider $(yx)^{n+1}=y(xy)^nx = yx$. This means that $(yx)^n=1$ and so $ord(yx)\le n=ord(xy)$. By symmetry, $ord(yx)\ge ord(xy)$ and so $ord(xy) = ord(yx)$.

Now apply this to $x=a_1$ and $y=a_2 \circ \cdots \circ a_n$.

Edit: A simpler proof that $ord(xy) = ord(yx)$ is to note that $xy$ and $yx$ and actually conjugate: $yx = z^{-1} (xy) z$, for $z=x$. And conjugate elements have the same order (the proof being along the same lines as above).

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I wouldn't really call the second one a 'simpler' proof. It's pretty much exactly as easy. – Tara B May 9 '12 at 12:08
@TaraB, Like I said, the proofs are quite similar, but I think the second one is conceptually easier. Opinions differ, of course. – lhf May 9 '12 at 12:10
I do think that proving that two conjugate elements having the same order is slightly simpler than showing $o(xy) = o(yx)$ 'from scratch', but then I find that having to observe that $xy$ and $yx$ are conjugate cancels out that slight advantage. Not that it's at all important! – Tara B May 9 '12 at 12:12
@TaraB: yes, but the observation that for any elements $x,y$ in any group $G$, $xy$ is conjugate to $yx$ is also very important, so having to realize it is (in my opinion, obviously) not really part of the "cost" of the proof. Think of how many other things you can prove knowing this: for instance, you're halfway to proving that for square matrices $A$ and $B$, $AB$ and $BA$ have the same characteristic polynomial. – Pete L. Clark May 9 '12 at 13:17
@user240718, you can cancel $yx$ in $(yx)^{n+1}= yx$. – lhf Jul 9 '15 at 12:55

I guess this is how I would pretend that the proof was by induction:

First do the $n=2$ case, as lhf has already done.

(I'm going to use $o(g)$ for the order of $g$, just because I think it looks nicer.)

Now suppose the statement is true for some $n\geq 2$ and let $a_1,a_2,\ldots,a_{n+1}\in G$. Then $a_n a_{n+1}\in G$, so $o(a_1 a_2 \ldots a_n a_{n+1} ) = o( a_1 a_2 \ldots (a_n a_{n+1}) ) = o( a_2 \ldots (a_n a_{n+1}) a_1) = o( a_2 \ldots a_n a_{n+1} a_1)$. Hence, by induction, the statement is true for all $n\in \mathbb{N}$.

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