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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!

big_GolfUniformIndia
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    Hint: count elements of order $4$ in both groups – Sofía Marlasca Aparicio Apr 30 '21 at 20:23
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    The image of the multiplication by $2$ homomorphism has 4 elements for the first group and only 2 for the second group. That "multiplication by $2$ homomorphism" would necessarily commute with any isomorphism, so that if they were indeed isomorphic the two images should have the same cardinality, which they don't. – Olivier Bégassat Apr 30 '21 at 20:28
  • @marlasca23 I've counted and $C_4 \times C_4$ has $12$ elements of order $4$, while $C_4 \times C_2 \times C_2$ has $8$, so they are not isomorphic. This was a way, but I thought that there was some easier way! Thank you for your help! – big_GolfUniformIndia Apr 30 '21 at 20:50

3 Answers3

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Count elements of order 2 in both groups.

Lorenzo Panebianco
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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.

Dietrich Burde
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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.

Shaun
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    I don't like to be critical, but that seems much more difficult than counting the number of elements of order $2$. – Derek Holt May 01 '21 at 09:00