Number of group homomorphisms from $Z_2\times Z_2\to Z_4$

Question

My reasoning is: there are four elements in $Z_2\times Z_2$, they are all divisors of $2$. This means that they must be sent to elements of order $2$ or $1$. There are only two elements of these orders in $Z_4$, plus the fact that we need the identity element $(0,0)\in Z_2\times Z_2$ to be sent to the identity $0\in Z_4$. So there are $2^3=8$ different homomorphisms.

Is this reasoning correct? How about ring homomorphisms?

Thanks.

Answer 1

Using kernels to find homomorphism,

As, normal subgroups of $\mathbb{Z}_{2} \times \mathbb{Z}_{2} $ are, $<(0,0)>,<(1,0)>,<(0,1)>,<(1,1)>,\mathbb{Z}_{2} \times \mathbb{Z}_{2} $.

For $<(0,0)>$ as kernel, there is no homomorphism .

For, $<(1,0)>,<(0,1)>,<(1,1)>$ as kernels, there are three non-trivial homomorphism , each one for each respective kernel.

For $\mathbb{Z}_{2} \times \mathbb{Z}_{2} $ as kernel, there is a trivial homomorphism.

If you are taking about ring homomorphism, then there is one trivial ring homomorphism , as $\bar{0}$ is idempotent in $\mathbb{Z}_{4} $

Answer 2

$G = C_2 \times C_2$

Let $a \in G$ generate the first $C_2$ and $b$ generate the second. Together $a,b$ generate all of $G$.

A group homomorphism $G \to C_4$ is completely determined by its action on $a$ and $b$ because

• $\phi(1) = 1$
• $\phi(ab) = \phi(a)\phi(b)$

Let $c$ generate $C_4$, as you said $\phi(a) = 1$ or $\phi(a) = c^2$ so there are 2 options for $a$ and 2 options for $b$.

In total 4 possible homomorphisms.