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Definition of an order of a group: Let $G$ be a group and $a\in G$. Then $a$ is said to have finite order if there exists a $n\ge 1$ such that $a^{n}=e$. If $a$ is of no finite order then $a$ is said to have infinite order.

The question states, let $a$ and $b$ be two elements of a group $G$ such that $a$ has order 3, b has order 2, and $bab^{-1}$=$a^{-1}$.
Prove that $ab$ had order 6 by showing none of $(ab)^{2}$,$(ab)^{3}$,...$(ab)^{5}$ are equal to $e$, but $(ab)^{6}$=$e$. Give an example of a group and elements $a$ and $b$ such that $a$ has order 3, $b$ has order 2, and $bab^{-1}$=$a^{-1}$.

I was only able to do $(ab)^{2}$. Here it is:
$(ab)^{2}$= $abab$
$\qquad$ =$abab^{-1}$
$\qquad$ =$a$ $\,$ $a^{-1}$=$1$

I am stuck trying to do the rest of this problem.

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    But then,this shows that $ab$ has order $2$,not $6$. – 2017-02-14
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    It says I have to show $ab$ has order 6 by showing none of $(ab)^{2}$=$e$. I'm trying figure this out myself – 2017-02-14
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    Yes, but the question is wrong, as the answer below shows. I know that it may seem unlikely, but that is the case. – 2017-02-14
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    Ok. I will notify my professor then – 2017-02-14
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    In the symmetric group $S_3$ let $a=(1\ 2\ 3)$ and $b=(1\ 2);$ then $a$ has order $3$ and $b$ has order $2$ and $bab^{-1}=(1\ 3\ 2)=a^{-1}.$ And of course $ab=(1\ 3)$ has order $2;$ there are no elements of order $6$ in $S_3.$ – 2017-02-14

2 Answers 2

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$bab^{-1}=a^{-1}$ implies that $aba=b$.We deduce that $(ab)^2=abab=b^2=e$. The order of $ab$ is 2 and not 6.

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    it says I have to show $ab$ has order 6 by showing none of $(ab)^{2}$=$e$. I'm trying figure that out myself. I see that your $bab^{-1}$= $a^{-1}$ solution looks about right. The question itself is weird to me – 2017-02-14
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    It's also kind of weird that the assumption that $a$ has order $3$ is not needed in showing $(ab)^2=e,$ and that the notation $b^{-1}$ is used when $b$ is of order $2.$ – 2017-02-14
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    exactly. I might just take the whatever I can get on the problem – 2017-02-14
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I'm going to outline what I think the problem should be, and solve that.

Let us suppose instead that $a^2=e$, $b^3=e$, along with $bab^{-1} = a^{-1} = a$. The latter is equivalent to $ab=ba$. Then $$\begin{align} (ab)^2 &= abab = baab = b^2 = b^{-1} \neq e \\ (ab)^3 &= (ab)b^{-1} = a \neq e \\ (ab)^4 &= ((ab)^2)^2 = b^4 = b \neq e \\ (ab)^5 &= abb = ab^2 = ab^{-1} \neq e \\ (ab)^6 &= ((ab)^3)^2 = a^2 = e \end{align} $$

Essentially, what happens here is that, because $a$ commutes with $b$, $(ab)^n = a^n b^n$, and you need $2\mid n$ for $a^n=e$ and $3 \mid n$ for $b^n=e$.

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    Okay. When I did $(ab)^{2}$ I got 1. For some reason you did your much different. This looks about right – 2017-02-14
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    @Chappers, You mixed up $a$ and $b$! It is $a$ that has order $3$ and $b$ has order $2$. – 2017-02-14
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    @NickyHekster As others have noted and I say in the first two sentences, the problem as posed is wrong. This is what I think the question wanted to ask: if one instead has the orders of $a$ and $b$ the other way round, $ab$ really does have order $6$. – 2017-02-14