3
$\begingroup$

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 $Z_4$ ⊕ $Z_4$.

$\endgroup$
  • $\begingroup$ 32>16, so perhaps you question might be rephrased $\endgroup$ – Thomas Dec 1 '15 at 9:57
  • $\begingroup$ You know wrong then! $\endgroup$ – Derek Holt Dec 1 '15 at 9:58
  • $\begingroup$ @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 $Z_5$ $\endgroup$ – Mehul Jain Dec 1 '15 at 10:09
  • $\begingroup$ 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 $\mathbb{Z}/m\mathbb{Z}$. $\endgroup$ – user60589 Dec 1 '15 at 10:43
  • $\begingroup$ @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 :) $\endgroup$ – Mehul Jain Dec 1 '15 at 11:05
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.