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Show that the group $U_8$ of units modulo $8$ is not isomorphic to $U_{10}$?

This is my answer. Check it for me please.

Suppose $U_8$ is isomorphic to $U_{10}$, $3 ∈ U_{10}$ and order of $3$ in $U_{10}$ is $4$. But there is no element of order $4$ in $U_8$, which is contradiction. $U_8$ is not isomorphic to $U_{10}$.

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seems fine to me –  Eli Elizirov Dec 8 '13 at 9:21

1 Answer 1

This is indeed correct. Note that isomorphisms will preserve the orders of elements.

Proof: Consider any $a \in G$ of order $n$, and suppose we have an isomorphism $\phi:G \rightarrow G'$. First, recall that, for any group homomorphism, the identity in the first group will always map to the identity in the second group. Thus:

$$\phi(e) = \phi(a^n) = \phi(a)^n = e$$

Since $3 \in U_{10}$ has order $4$ and there is no element of order $4$ in $U_8$, then we conclude $U_{10} \ncong U_8$.

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