Finding the maximum order of an element of $U(4) \times U(9) \times U(5)$

Question

I would like to find the maximum order of an element of $U(4) \times U(9) \times U(5)$. I know that $U(4) = U(2^2)$, and $U(9) = U(3^2)$. So I get that $U(4) \times U(9) \times U(5) = U(2^2) \times U(3^2) \times U(5).$

I also know that $U(2^2) \cong \mathbb Z_2 \oplus\mathbb Z$ and $U(3^9) \cong \mathbb Z_6$. So we get that $U(2^2) \times U(3^2) \times U(5) \cong (\mathbb Z_2 \oplus\mathbb Z) \oplus \mathbb Z_6 \oplus \mathbb Z_4$.

Clearly, the maximum order of an element of $U(4) \times U(9) \times U(5)$ is finite. But one of the isomorphism theorems I've applied above tells me that that particular maximum order of an element of $U(4) \times U(9) \times U(5)$ is equal to the maximum order of an element of $(\mathbb Z_2 \oplus\mathbb Z) \oplus \mathbb Z_6 \oplus \mathbb Z_4$, which is, in fact, infinite. I don't get this at all.

$(\mathbb Z_2 \oplus\mathbb Z) \oplus \mathbb Z_6 \oplus \mathbb Z_4$ would have infinitely many elements since $\mathbb Z_2 \oplus\mathbb Z$ has infinitely many elements.

Can someone please explain this to me?

Thanks in advance.

Answer 1

$U(4)\cong\Bbb Z_2$.

We wind up with $\Bbb Z_2×\Bbb Z_6×\Bbb Z_4\cong\Bbb Z_2×(\Bbb Z_2×\Bbb Z_3)×\Bbb Z_4\cong\Bbb Z_2×\Bbb Z_2×(\Bbb Z_3×\Bbb Z_4)\cong\Bbb Z_2×\Bbb Z_2×\Bbb Z_{12}$.

There is thus an element of order twelve.

Answer 2

This is an issue of notation.

Remember that $\mathbb{Z}_{n}$ actually means the quotient group $\mathbb{Z} / n\mathbb{Z}$, i.e., the group of integer remainders modulo $n$.

So $\mathbb{Z}_1 \cong \mathbb{Z} / 1\mathbb{Z} \cong \{0\}$, the trivial group. The only possible remainder when dividing by 1 is 0. So $\mathbb{Z}_1 \ncong \mathbb{Z}$! A reasonable mistake to make!