Direct group isomorphism, order. Show that $\def\Z{\mathbb{Z}}\Z_4 \times \Z_4$ is not isomorphic to $\Z_4 \times \Z_2 \times \Z_2$. (Hint: count elements of order 4 in the groups).
Attempt: I have tried to find the order of each pair for $\Z_4 \times \Z_4$.
So, $|(0,1)| = 4$, $|(0,3)| = 4$, $|(1,3)|=4$, $|(3,0)| = 4$, $|(3,1)| = 4$, $|(3,2)| = 4$, $|(3,4)|=4$, are pairs with order $4$. Then if I am not mistaken, all other pairs in $\Z_4 \times \Z_4$ have order $2$.
Can anyone please help for the order of some elements in $\Z_4 \times \Z_2 \times \Z_2$? I think if $\Z_4 \times \Z_4$ has elements of orders $2$ and $4$, then $\Z_4 \times \Z_2 \times \Z_2$ also has to have orders $2$ and $4$ for it to be isomorphic. 
Thank you very much.
 A: It suffice to show $\mathbb Z_2 \times \mathbb Z_2$ is not isomorphic to $\mathbb Z_4$. However this is quite simple, since no elements in $\mathbb Z_2 \times \mathbb Z_2$ has order $4$, however $1 \in \mathbb Z_4$ clearly has order $4$.
You could also count elements of order $2$ in group $G_1= \mathbb Z_4 \times \mathbb Z_2 \times \mathbb Z_2$ directly.
A: Each copy of $\mathbb{Z}_4$ has two elements of order $4$, namely $1$ and $3$. Note that the order of any element $(a, b)$ will be $lcm(|a|, |b|)$.  Hence, we can get $10$ elements of order $4$ in $\mathbb{Z}_4 \times \mathbb{Z}_4$.
$$(1, 0); (0, 1)$$
$$(3, 0); (0, 3)$$
$$(1, 1)$$
$$(3, 3)$$
$$(2, 1); (1, 2)$$
$$(2, 3); (3, 2)$$
Now let's take a look at $\mathbb{Z_4} \times \mathbb{Z}_2 \times \mathbb{Z}_2$.
Again, note that the order of any element $(a, b, c)$ will be $lcm(|a|, |b|, |c|)$.  Therefore, the order of any element $(a, b, c)$ will simply be $|a|$.  There are only $8$ ways to pair an order $4$ element with an element from $\mathbb{Z}_2 \times \mathbb{Z}_2$.  Thus, there are only $8$ order $4$ elements in this group.
