element

**Sampras** · November 9th 2009, 01:38 AM

Suppose $a,b \in G$ and $ab = ba$ . If $|a|$ and $|b|$ are relatively prime, prove that $ab$ has order $|a||b|$ .

So suppose $a^m = e$ and $b^n = e$ . Then $\gcd(m,n) = 1$ . Thus $1 = \frac{\text{lcm}(m,n)}{mn}$ or $mn = \text{lcm}(m,n)$ . So $(ab)^{mn} = (a^{m})^{n} \cdot (b^{n})^{m} = e^{n}e^{m} = e$ . Thus the order of $ab$ is $|a||b|$ .

Is this correct?

**Drexel28** · November 9th 2009, 03:24 AM

Originally Posted by Sampras

Suppose $a,b \in G$ and $ab = ba$ . If $|a|$ and $|b|$ are relatively prime, prove that $ab$ has order $|a||b|$ .

So suppose $a^m = e$ and $b^n = e$ . Then $\gcd(m,n) = 1$ . Thus $1 = \frac{\text{lcm}(m,n)}{mn}$ or $mn = \text{lcm}(m,n)$ . So $(ab)^{mn} = (a^{m})^{n} \cdot (b^{n})^{m} = e^{n}e^{m} = e$ . Thus the order of $ab$ is $|a||b|$ .

Is this correct?

I suppose that is fine. Are you allowed to use that result? It makes the proof kind of trivial.

**Sampras** · November 9th 2009, 03:53 AM

Originally Posted by Drexel28

I suppose that is fine. Are you allowed to use that result? It makes the proof kind of trivial.

What would you suggest?

**tonio** · November 9th 2009, 04:06 AM

Originally Posted by Sampras

Suppose $a,b \in G$ and $ab = ba$ . If $|a|$ and $|b|$ are relatively prime, prove that $ab$ has order $|a||b|$ .

So suppose $a^m = e$ and $b^n = e$ . Then $\gcd(m,n) = 1$ . Thus $1 = \frac{\text{lcm}(m,n)}{mn}$ or $mn = \text{lcm}(m,n)$ . So $(ab)^{mn} = (a^{m})^{n} \cdot (b^{n})^{m} = e^{n}e^{m} = e$ . Thus the order of $ab$ is $|a||b|$ .

Is this correct?

I'm afraid it's not: you only proved that $(ab)^{mn}=e$ which is pretty trivial. The interesting part is in showing that there is not power of ab less than mn which equals the unit...

Tonio

**Sampras** · November 9th 2009, 04:18 AM

Originally Posted by tonio

I'm afraid it's not: you only proved that $(ab)^{mn}=e$ which is pretty trivial. The interesting part is in showing that there is not power of ab less than mn which equals the unit...

Tonio

You have to use fact that ab=ba, because this isn't true if this condition is not met? So $(ab)^{mn} = (ba)^{mn}$ . Suppose $(ab)^{mn- \varepsilon} = e = (ba)^{mn- \varepsilon}$ . Then $\frac{(ab)^{mn}}{(ab)^{\varepsilon}} = e = \frac{(ba)^{mn}}{(ba)^{\varepsilon}}$ for some $\varepsilon \in \mathbb{Z}$ .

Or $\frac{a^{mn}b^{mn}}{a^{\varepsilon}b^{\varepsilon} } = e$ . Or $\frac{e}{a^{\varepsilon} b^{\varepsilon}} = e$ so that $(ab)^{\varepsilon} = e$ . Contradiction?

**Drexel28** · November 9th 2009, 06:24 AM

NOTE: I am actually not sure if this is correct. Let another member validate this. I feel as though I may have made a stupid incorrect assumption.

Problem: Let $(m,n)=1$ . Prove that if $|a|=m,|b|=n$ that $|ab|=mn$ .

Proof: Clearly $\left(ab\right)^{mn}=e$ . Now suppose that $|ab|=\omega<mn$ then $(ab)^{\omega}=e\implies a^{\omega}=b^{-\omega}$ and $b^{\omega}=a^{-\omega}$ . Thus $\left(b^{-\omega}\right)^m=\left(a^{\omega}\right)^m=e\impli es a^{\omega\cdot m-\omega\cdot n}=e$ . Similarly $b^{\omega\cdot n-\omega\cdot m}=e$ . And since $|a|=m$ we see that $m|\omega\cdot m-\omega\cdot n\implies m|\omega n$ and since $(m,n)=1$ this is only true if $m|\omega$ . Similarly $n|\omega\cdot n-\omega\cdot m$ and since $|b|=n$ this means $n|\omega\cdot n-\omega\cdot m\implies n|\omega\cdot m\$ and by previous reasoning $n|\omega$ . Therefore $m,n|\omega$ which means that $\text{lcm}(m,n)|\omega$ but since $(m,n)=1$ we have that $\text{lcm}(m,n)=mn$ and $mn|\omega$ which is a contradiction since $0<\omega<mn$ .

**Sampras** · November 9th 2009, 06:33 AM

Originally Posted by Drexel28

NOTE: I am actually not sure if this is correct. Let another member validate this. I feel as though I may have made a stupid incorrect assumption.

Problem: Let $(m,n)=1$ . Prove that if $|a|=m,|b|=n$ that $|ab|=mn$ .

Proof: Clearly $\left(ab\right)^{mn}=e$ . Now suppose that $|ab|=\omega<mn$ then $(ab)^{\omega}=e\implies a^{\omega}=b^{-\omega}$ and $b^{\omega}=a^{-\omega}$ . Thus $\left(b^{-\omega}\right)^m=\left(a^{\omega}\right)^m=e\impli es a^{\omega\cdot m-\omega\cdot n}=e$ . Similarly $b^{\omega\cdot n-\omega\cdot m}=e$ . And since $|a|=m$ we see that $m|\omega\cdot m-\omega\cdot n\implies m|\omega n$ and since $(m,n)=1$ this is only true if $m|\omega$ . Similarly $n|\omega\cdot n-\omega\cdot m$ and since $|b|=n$ this means $n|\omega\cdot n-\omega\cdot m\implies n|\omega\cdot m\$ and by previous reasoning $n|\omega$ . Therefore $m,n|\omega$ which means that $\text{lcm}(m,n)|\omega$ but since $(m,n)=1$ we have that $\text{lcm}(m,n)=mn$ and $mn|\omega$ which is a contradiction since $0<\omega<mn$ .

Suppose $a^{r}b^{r} = e$ for some $r \in \mathbb{Z}^{+}$ . Then $b^{nr-r}b^{r} = b^{nr} = e$ . Let $t = nr-r$ . Then $a^{r} = b^t$ . So $(a^r)^n = (b^t)^n$ so that $a^{rn} = e$ . So $m$ divides $rn$ but $(m,n) = 1$ . Thus $m|r$ so that $m$ divides $|a|b|$ . Similarly, $n$ divides $|a|b|$ .

**Drexel28** · November 9th 2009, 06:34 AM

Originally Posted by Sampras

Suppose $a^{r}b^{r} = e$ for some $r \in \mathbb{Z}^{+}$ . Then $b^{nr-r}b^{r} = b^{nr} = e$ . Let $t = nr-r$ . Then $a^{r} = b^t$ . So $(a^r)^n = (b^t)^n$ so that $a^{rn} = e$ . So $m$ divides $rn$ but $(m,n) = 1$ . Thus $m|r$ so that $m|ab$ . Similarly $n|ab$ .

So wait haha. Are you agreeing or disagreeing with me?

**Sampras** · November 9th 2009, 06:39 AM

Originally Posted by Drexel28

So wait haha. Are you agreeing or disagreeing with me?

It's essentially the same proof. Except just choosing an arbitrary r to begin with.

**Drexel28** · November 9th 2009, 06:40 AM

Originally Posted by Sampras

It's essentially the same proof. Except just choosing an arbitrary r to begin with.

Ok. Good. Hope that helped.

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