Why is $U(8)$ isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2$?

Question

I understand that all the elements of $U(8)$ all have order $2$. However, why is it that $U(8)$ is isomorphic to $\mathbb{Z}_2\oplus\mathbb{Z}_2$ and not just $\mathbb{Z}_2$?

Answer 1

Because $\mathbb{Z}_2$ only has 2 elements. While $U_8$ has 4 elements. Thus the groups have a different order, and therefore can not be isomorphic.

The order of a group is different from the order of an element of the group. The order of a group is simply the size of the group, while the order of an element is the $n$ where $a^n = e$ with $a$ being an element of the group.

For two groups to be isomorphic their elements have to have the same order. For example $\mathbb{Z}_4 \not\cong \mathbb{Z}_2 \times \mathbb{Z}_2$ because $\mathbb{Z}_4$ has an element of order 4, while $\mathbb{Z}_2 \times \mathbb{Z}_2$ does not, even tho the order of the two groups are the same.