3

Let $G$ and $H$ be the abelian groups $G=\mathbb{Z}/30\mathbb{Z}\oplus\mathbb{Z}$ and $H=\mathbb{Z}/15\mathbb{Z}\oplus\mathbb{Z}/7\mathbb{Z}$. Determine the number of group homomorphisms from $G$ to $H$, that is, the number of elements of $Hom_{\mathbb{Z}}(G,H)$.

How do I find the mappings that could keep the operations from $G$ to $H$? Can I start to do it by separating the problem into $\mathbb{Z}/30\mathbb{Z}\to\mathbb{Z}/15\mathbb{Z}$ and $\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}$, and multiply them together?

Thank you very much!

breezeintopl
  • 1,437

1 Answers1

5

Hint: Let $A_i, 1 \leq i \leq n$ be a finite collection of abelian groups and let $B$ be another abelian group. Then $Hom(\oplus_{i=1}^n A_i, B)\cong \oplus_{i=1}^n Hom(A_i, B)$ and $Hom(B, \oplus_{i=1}^n A_i)\cong \oplus_{i=1}^n Hom(B, A_i).$

Using this we have $$Hom(G,H) \cong Hom(\mathbb Z/30\mathbb Z, \mathbb Z/15\mathbb Z) \oplus Hom(\mathbb Z/30\mathbb Z, \mathbb Z/7\mathbb Z) \oplus Hom(\mathbb Z, \mathbb Z/15\mathbb Z) \oplus Hom(\mathbb Z, \mathbb Z/7\mathbb Z).$$

Added: (1). To define a group homomorpshism $\phi: G_1 \to G_2,$ where $G_1$ is a cyclic group, it's enough to define on a generator of $G_1.$ In general, $Hom(\mathbb Z, m\mathbb Z, \mathbb Z/n\mathbb Z) \cong \mathbb Z/d\mathbb Z,$ where $d=gcd(m,n).$

(2). To find a homomorphism $\phi: \mathbb Z/30\mathbb Z \to \mathbb Z/15\mathbb Z,$ we need to find the possible images of $\bar 1$ (as noted above). Also note that, $|\phi (\bar 1)|$ must divide 30. Using the same argument it's easy to show that the only homomorphism $\mathbb Z/30\mathbb Z \to \mathbb Z/7\mathbb Z$ is the zero homomorphism.

Krish
  • 7,102
  • Thank you so much for your answer! For $Hom(\mathbb{Z},\mathbb{F}{15})$, I think $\phi(x)=0$, $\phi(x)=x\quad mod\quad15$ is in it. Similar to $Hom(\mathbb{Z},\mathbb{F}{7})$. For $Hom(\mathbb{F}{30},\mathbb{F}{7})$, only $\phi(x)=0$ is in it? For $Hom(\mathbb{F}{30},\mathbb{F}{15})$, $\phi(x)=0$, $\phi(x)=x\quad mod\quad 15$ is in it? So generally, how can we determine $Hom(\mathbb{F}{m},\mathbb{F}{n})$? – breezeintopl Mar 20 '15 at 20:47
  • @breezeintopl: I've added an explanation regarding your comment. I hope this will answer your questions. – Krish Mar 21 '15 at 14:00
  • Thank you so much for your answer! So for $Hom(\mathbb{Z}, \mathbb{Z}/n\mathbb{Z})$, it should isomorphism to $\mathbb{Z}/n\mathbb{Z}$, right? – breezeintopl Mar 23 '15 at 19:37
  • Yes! That's true. – Krish Mar 24 '15 at 06:18
  • Is the result NOT TRUE for non-abelian groups ? @ Krish Please tell me...I am confused about this question after seeing answer of this question. I am unable to understand the difference between that answer and your answer.... – Empty Apr 11 '15 at 12:43
  • @S.Panja-1729: That link shows that the relation $Hom(B, \oplus_{i=1}^n A_i)\cong \oplus_{i=1}^n Hom(B, A_i)$ is true without the assumption "abelian". But here we use two relations and the first one, I think, is not true in general. Try $A_1=A_2=\mathbb Z/2\mathbb Z, B=S_3.$ – Krish May 01 '15 at 05:15