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Let $G$ be a group, denote for $g \in G:$ $\langle g \rangle=\{g^0,g^1,\ldots \}$

I should show that for $x,y \in G$ with $xy=yx$ if $ord(\langle x \rangle)=l<\infty$ and $ord(\langle y \rangle)=m<\infty$ then $ord(\langle xy \rangle)<\infty$

To show it I assumed $ord(\langle xy \rangle)=\infty$, then especially

$$((xy)^l)^m=(x^l)^m\cdot (y^m)^l=e^me^l=e\neq e$$

But that is a contradiction. However I did not use that $xy=yx$ so the proof should be incorrect, but where is my mistake and how to show it correctly?

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You actually used it. Otherwise it would be false that $(xy)^a=x^a\cdot y^a$! For instance, in general $(xy)^2=(xy) \cdot (xy)=xyxy \neq xxyy=x^2y^2$ unless $xy=yx$. Also you need not reason by contradiction. You can simply say that under the commutation hypothesis $(xy)^{ml}=x^{ml}\cdot y^{ml}=(x^l)^m\cdot (y^m)^l=e^m\cdot e^l=e$ thus showing directly that $xy$ has finite order $\leq ml$. –  Olivier Bégassat Oct 31 '11 at 19:42
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Yes, thank you *facepalm* :-) –  Listing Oct 31 '11 at 19:48
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2 Answers 2

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You used the fact that $xy=yx$ when you split up exponents over $xy$ - in general, for any $d>1$, $$(xy)^d=\underbrace{(xy)(xy)\cdots(xy)}_{d\text{ times}}\neq \left(\underbrace{ x\cdot x\cdots x}_{d\text{ times}}\right)\left(\underbrace{ y\cdot y\cdots y}_{d\text{ times}}\right)=x^dy^d$$ unless you can turn the successive $yx$'s on the left into $xy$'s: $$\underbrace{(xy)(xy)\cdots (xy)}_{d\text{ times}}=x\underbrace{(yx)(yx)\cdots(yx)}_{(d-1)\text{ times}}y=x\underbrace{(xy)(xy)\cdots(xy)}_{(d-1)\text{ times}}y=x^2\underbrace{(yx)(yx)\cdots(yx)}_{(d-2)\text{ times}}y^2$$ $$(\text{ repeat })\cdots = x^dy^d$$


Also, note that you did not need to use a proof by contradiction - you have directly proven that $$(xy)^{lm}=x^{lm}y^{lm}=(x^l)^m(y^m)^l=e^me^l=ee=e$$ so there was no need to assume $\text{ord}(xy)=\infty$ and then show it was false. It is not wrong, but it is much cleaner to say

Here is a proof of claim $P$.

than to say

Suppose claim $P$ were false. But here is a proof that it is true; contradiction. Thus, our assumption that claim $P$ is false must have been false, i.e. the claim $P$ is true.

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You actually did use that property when you said $$((xy)^l)^m=(x^l)^m⋅(y^m)^l$$ since, for example $$(xy)^3=xyxyxy$$ which is not necessarily the same as $xxxyyy$ if $x$ and $y$ do not commute.

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