# Number of group homomorphism from $Z_8$ ⊕ $Z_2$ to $Z_4$ ⊕ $Z_4$

I know that there does not exist an isomorphism from $Z_8$ ⊕ $Z_2$ to $Z_4$ ⊕ $Z_4$ as there exist an element of order 8 in $Z_8$ ⊕ $Z_2$ and no element of order 8 in $Z_4$ ⊕ $Z_4$.
But what about homomorphisms. How many of them exist and how to find them.

There does exist homomorphisms. Example: if (1,0) of $Z_8$ ⊕ $Z_2$ is mapped to (1,0) of $Z_4$ ⊕ $Z_4$ and (0,1) of $Z_8$ ⊕ $Z_2$ mapped to (0,0) of $Z_4$ ⊕ $Z_4$.

• 32>16, so perhaps you question might be rephrased – Thomas Dec 1 '15 at 9:57
• You know wrong then! – Derek Holt Dec 1 '15 at 9:58
• @Thomas . I made a typo earlier but 32>16 does not ensure that there does not exist a homomorphism. For example there exist 5 homomorphisms from $Z$ to $Z_5$ – Mehul Jain Dec 1 '15 at 10:09
• The direct sum is a biproduct, so $\text{Hom}(A \oplus B, C) \cong \text{Hom}(A, C) \times \text{Hom}(B, C)$ and $\text{Hom}(A , B \oplus C) \cong \text{Hom}(A , B \times C) \cong \text{Hom}(A, B) \times \text{Hom}(A, C)$. So you only need the number of homomorphisms from $\mathbb{Z}/n\mathbb{Z}$ to $\mathbb{Z}/m\mathbb{Z}$. – user60589 Dec 1 '15 at 10:43
• @user60589 Great!! I got the correct answer by the above mentioned property of bi-products. The answer comes out to be $2^6$ which is correct. It will be nice of you if you write an elaborate answer of this for anyone who encounter the problem in future. Though I am still thinking of how to prove the above mentioned property. It look interesting. Thanks :) – Mehul Jain Dec 1 '15 at 11:05

The direct sum is a biproduct in the category of abelian groups. This means that is the coproduct and the product at the same time.

The coproduct $A\coprod B$ has the universal property that any map $f\colon A\coprod B\to C$ is uniquely defined by the two maps $f\circ i_1$ and $f\circ i_2$ where $i_1$ (resp. $i_2$) is the inclusion of $A$ (resp. $B$) in $A\coprod B$.

Using this you can show that $$\text{Hom}\left(A\coprod B, C\right) \cong \text{Hom}(A, C) \times \text{Hom}(B, C) .$$

Dually, (it is the same with domain and codomain exchanged) a map into the product is uniquely determined by the composition with the projections of the product. So you have $$\text{Hom}(A, B \times C) \cong \text{Hom}(A, B ) \times \text{Hom}(A, C) .$$

Now we know that $$\text{Hom}(\mathbb{Z}/8\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}) \cong \text{Hom}(\mathbb{Z}/8\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}) \times \text{Hom}(\mathbb{Z}/8\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}) \times \text{Hom}( \mathbb{Z}/2\mathbb{Z}, \mathbb{Z}/4\mathbb{Z}) \times \text{Hom}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}) .$$

Using $$\text{Hom}(\mathbb{Z}/ n\mathbb{Z}, \mathbb{Z}/ m\mathbb{Z}) \cong \mathbb{Z} / gcd(n,m) \mathbb{Z}$$ it simplifies to $$\text{Hom}(\mathbb{Z}/8\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z},\mathbb{Z}/4\mathbb{Z}\oplus \mathbb{Z}/4\mathbb{Z}) \cong \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z} \times \mathbb{Z}/2 \mathbb{Z} .$$

Thus there are $2^6$ homomorphisms.