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Problem 68 (The Orders Of Elements Match In Isomorphic Groups)
Suppose that $f:G\to H$ a group isomorphism. Suppose that $f(g)=h$. Prove that the order of $g$ and the order of $h$ are the same, in other words show that $|g|=|h|$. In particular, since the only elements of order 1 in each group are the identities $e_G\in G$ and $e_H\in H$, then we must have $f(e_G)=e_H$.
Click here to see a partial proof that's missing one piece.
Suppose that $f:G\to H$ is an isomorphism and that $e_G$ and $e_H$ are the identities in $G$ and $H$ respectively. We first show that $f(e_G)=e_H$. Note that $e_Ge_G=e_G$. Apply $f$ to both sides to obtain $f(e_G)f(e_G)=f(e_G)$. Since $f(e_G)=f(e_G)e_H$, we now have $f(e_G)f(e_G)=f(e_G)\cdot e_H.$ Cancelling $f(e_G)$ from the left of both sides gives $f(e_G)=e_H$ as desired.
We now prove that if $a$ has order $n$, then $f(a)$ must have order $n$ as well. Suppose $a\in G$ has order $n$. This means that $n$ is the smallest positive integer such that $a^n=e_G$, or that $aa \cdots a=e$ (repeated $n$ times). If we apply $f$ to both sides of this equation, then we see that $f(a)f(a)\cdots f(a)=f(e_G)$. This means that $f(a)^n=f(e_G)$. Since we know $f(e_G)=e_H$, then we have shown $f(a)^n=e_H$ is the identity. This means the order of $f(a)$ must equal $n$. $\square$
What's missing? Pay close attention to the definition of order.
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