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Suppose that $G_1$ and $G_2$ are finite groups and $\beta: G_1\to G_2$ is an isomorphism. If $x_2 = \beta(x_1)$ for a given element $x_1 \in G_1$, prove that $x_1$ and $x_2$ have the same order.

I think I should use induction but I am not sure how could I prove it?

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    Show $\beta(x_1^n) = \beta(x_1)^n$, possibly via induction... – 2017-01-18
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    What should be the base case and induction assumption?? – 2017-01-18
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    Related: http://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 . Note: your title asks about proving (something is) an isomorphism. Your question is about the consequences of having an isomorphism. – 2017-01-18
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    I leave that to you, but it might be good to start with the very definition of a homomorphism. – 2017-01-18

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You don't need induction: saying $x_1$ has order $n$ means there's a isomorphism $\mathbf Z/n\mathbf Z\longrightarrow \langle x_1\rangle$ (mapping $1$ to $x_1$). Just left-compose this isomorphism with $\beta$ to obtain an isomorphism $\mathbf Z/n\mathbf Z\longrightarrow \langle x_2\rangle$, mapping $1$ to $x_2$.