Why is $C_4 \times C_4$ not isomorphic to $C_4 \times C_2 \times C_2$?

Question

a while ago I was trying to prove this:

Show that $C_4 \times C_4$ is not isomorphic to $C_4 \times C_2 \times C_2$.

I know that we can write $C_4 \times C_2 \times C_2$ as $C_4 \times V_4$, where $V_4$ is the $4$-Klein group, but I can't conclude the proof, because they are both abelian. But, is easy to show that $C_4$ is not isomorphic to $C_2 \times C_2$ ($C_4$ has an element of order $4$ and $C_2 \times C_2$ hasn't). Is it enough? I mean, is true that $A \ncong B \implies C \times A \ncong C \times B$?

Any help would be very appreciated! Thanks in advance!

Answer 1

Addressing your question "I mean, is true that $A \ncong B \implies C \times A \ncong C \times B$?":

Yes, finite abelian groups are cancellable, see this post:

For groups $A,B,C$, if $A\times B$ and $A\times C$ are isomorphic do we have $B$ isomorphic to $C$?

So $A\times B\cong A\times C$ implies that $B\cong C$. So we would obtain $C_4\cong C_2\times C_2$, which is a contradiction, since one group is cyclic, the other not.

Answer 2

Count elements of order 2 in both groups.

Answer 3

The group $\Bbb Z_4\times \Bbb Z_4$ has $15$ subgroups, whereas $\Bbb Z_4\times \Bbb Z_2\times \Bbb Z_2$ has $27$.

Isomorphisms preserve the number of subgroups.