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Order of a group’s element

Let \mathcal{G} be a group and a, b \in \mathcal{G} such that a^5=e and

(1)   \begin{equation*} aba^{-1}=b^2  \end{equation*}

where e is the identity element of the group. Find the order of b.

Solution

We will begin stating a lemma:

Lemma: If aba^{-1}=b^r then a^n b a^{-n}=b^{r^n}.

Proof:


First we multiply with a and a^{-1} from right and left respectively. Thus one can see that

(2)   \begin{equation*}a^2ba^{−2}=ab^ra^{−1}  \end{equation*}

Thus

(3)   \begin{equation*} ab^ra^{−1}=(aba^{−1})^r \end{equation*}

and

(4)   \begin{equation*} a^2ba^{−2}=(aba^{−1})^r \end{equation*}

Now, we use the main relation and so

a^2ba^{−2}=b^{r^2}

By repeating the previous procedure, one can prove the result.

Using the lemma we see that a^nba^{-n}=b^{2^n} and thus b^{31}=e. Since 31 is prime the order of b will be either 31 or 1.

 

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