For all subgroups $K \leq U(65)$, $K \ncong \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$.

Question

How would I show that for all subgroups $K \leq U(65)$, $K \ncong \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$?

I know that $U(65) \cong \mathbb Z_4 \oplus \mathbb Z_{12}$, in which $\mathbb Z_4 \oplus \mathbb Z_{12}$ has three elements of order 2 and $\mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$ has seven elements of order 2. How does that deduce that $K \ncong \mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_2$ for all subgroups $K$ of $U(65)$?

This is just what I'm reading from an answer text, which doesn't make much sense.

Thanks in advance.

Answer 1

The point being made is this: consider a fixed subgroup $K \subset G$. If $K \cong \Bbb Z_2 \oplus \Bbb Z_2 \oplus \Bbb Z_2$, then $K$ must also have seven elements of order $2$. However, since $K$ is a subset of $U(65)$, $K$ can have at most $3$ elements of order $2$.

Answer 2

Exercise. If $A$ and $B$ are isomorphic groups then the number of elements of $A$ of order two is the same as the number of elements of $B$ of order two.

Important Note: If the exercise is not obvious you haven't quite got what "isomorphic" means. Anything you say about $A$ ("in the language of group theory") that's true of $A$ is also true of $B$, because they are in some sense the same group! Maybe $B$ consists of the elements of $A$ painted black...

Hint: Say $f:A\to B$ is an isomorphism. Show that $x\in A$ has order two if and only if $f(x)\in B$ has order two.