Prove ord(a) = ord (bab(adsbygoogle = window.adsbygoogle || []).push({}); ^{-1}).

Attempt at the proof.

Proof:

Let G be a group and let a,b ε G.

(=>)

Suppose ord(a)=n. That is, a^{n}=e. We wish to that the ord(bab^{-1})=n. If we take a^{n}=e and multiply on the left by b and on the right by b^{-1}then, we have,

(1) a^{n}=e

(2) ba^{n}b^{-1}=beb^{-1}

(3) ba^{n}b^{-1}=e.

Hence, we have the fact that,

(4) a^{n}=e=ba^{n}b^{-1}.

Now, if we expand (bab^{-1})^{n}, we get,

(5) bab^{-1}bab^{-1}bab^{-1}..bab^{-1}=ba^{n}b^{-1}.

But, ba^{n}b^{-1}= e by our algebra in (1) - (4). Thus, (bab^{-1})^{n}=ba^{n}b^{-1}= e and ord(bab^{-1})=n.

(<=)

Let ord(bab^{-1})=n. That is, (bab^{-1})^{n}=e. But, as it was shown in (5), (bab^{-1})^{n}=ba^{n}b^{-1}= e. Using the fact that ba^{n}b^{-1}= e, we multiply on the left by b^{-1}and on the right by b and obtain that a^{n}=e. Hence, ord(a)=n.

QED

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# Homework Help: Proof about orders of elements. Ord(a)=ord(bab^-1).

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