By comparing orders of elements , we need to prove that $Z_{8} \times Z_{2}$ is not isomorphic to $Z_{4} \times Z_{4}$.
What I tried :
I observed number of elements of order $4$ in both the groups ,
$Z_{8} \times Z_{2}$ has $3$ elements with order $4$ and $Z_{4} \times Z_{4}$ has 5 elements with order $4$.
In isomorphism both the groups have same number of elements of same order .. Am I right for saying this ?
Is the above proof OK ?
Best Answer
Clearly this is the good idea. However, you made some mistakes. The number of element of order $4$ in $Z_8\times Z_2$ is $4$. The number of elements of order $4$ in $Z_4\times Z_4$ is $12$.
BTW the number of elements of order $8$ seems to be easier to deal with in this case.