Question

Q.Prove that two conjugate groups have same order.

Answer 1

Answer on Question #60412 – Math – Abstract Algebra

Question

Prove that two conjugates have the same order.

Proof

We need to prove that $g$ and $xgx^{-1}$ have the same order. It follows from the formula

$$(xgx^{-1})^n = xg^n x^{-1},$$

which shows $(xgx^{-1})^n = 1$ if and only if $g^n = 1$.

To see this, $(xgx^{-1})^n = 1$ implies

$$\underbrace{xgx^{-1} \cdot xgx^{-1} \cdot xgx^{-1} \cdots xgx^{-1}xgx^{-1}}_{n} = xg(x^{-1}x)g(x^{-1}x)g \cdots g(x^{-1}x)gx^{-1} = xg1g1g \cdots$$ $$g1gx^{-1} = xg \cdot g \cdot g \cdots g \cdot g \cdot x^{-1} = xg^n x^{-1} = 1,$$

hence $xg^n = x$,

that is, $g^n = 1$ by left-multiplying by $x^{-1}$.

The other direction of proof is clear.

It follows from this that the order of $g$ divides the order of $xgx^{-1}$ and vice versa, so the orders of $g$ and $xgx^{-1}$ must be equal.

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