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(a) Let $p$ be a prime. Determine the number of homomorphisms from $\Bbb Z_p \oplus \Bbb Z_p$ into $\Bbb Z_p$.

Attempt: Suppose $\Psi:Z_p \oplus Z_p \rightarrow Z_p$ is an into homomorphism.

Then $|Ker~\Psi|= p^2/x $ when $x = |\Psi (Z_p \oplus Z_p)| < p$

The only $p^2/x $ is an integer when $x=1$. Hence, there exists only the trivial homomorphism when $|\Psi (Z_p \oplus Z_p)| < p$ .

IS this correct?

Now, if we are asked the number of onto homomorphisms from $Z_p \oplus Z_p$ onto $Z_p$?

Attempt: If $\Psi:Z_p \oplus Z_p \rightarrow Z_p$ is an onto homomorphism, then $|Ker~\Psi|=p$

Since $Z_p$ is a cyclic (hence, normal) group, thus, $\Psi^{-1}(Z_p)$ is also a normal subgroup of $Z_p \oplus Z_p$ . (Anyways, every subgroup of an abelian group is normal)

I really am not able to move forward now.

(b) Determine all homomorphisms from $\Bbb Z$ onto $S_3$.

Attempt: $\Psi(1)$ can take $6$ values since $|S_3|=6$ . Hence, there are 6 possible homomorphisms ?

But, let $\Psi : Z \rightarrow S_3$ be a homomorphism. Since, the kernel of a homomorphism is a normal subgroup of the domain, and every subgroup of $Z$ is of the form $nZ \implies $ kernel of $Z$ must be of the form $nZ$ for some $n$. But $Z/nZ \approx Z_n$ and by the first isomorphism theorem : $Z/nZ$ should be isomorphic to $S_3$ but $S_3$ is not cyclic. So, how can this homomorphism hold true?

Now, we got to find all homomorphisms from $Z \rightarrow S_3$

Attempt: Since, $|S_3|=6$ , this means since the divisors of $6$ are $2,3$, we need to find cyclic subgroups of orders $2$ and $3$ in $S_3$. Let them be $H_2$ and $H_3$ respectively.

Then $Z/2Z \approx H_2$ and $Z/3Z \approx H_3$

IS this approach correct?

(c) Suppose that the number of homomorphisms from $G \mapsto H$ is $n$. If $H$ is abelian, how many homomorphisms are there from $G \oplus G\cdot\cdot \cdot \oplus~ G ( s $ times) to $H$?

I have really No idea how to proceed in this.

I have posted this question earlier but the book which I am reading (Gallian) has not introduced any such concepts as given in the answer.

Your help will be really appreciated. Thank you for your time and patience.

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I'm afraid your approach is not good.


The direct sum of abelian groups has a very important property.

Let $A$ and $B$ be abelian group. Then, consider the maps $i_A\colon A\to A\oplus B$, where $i_A(a)=(a,0)$, and $i_B\colon B\to A\oplus B$, where $i_B(b)=(0,b)$. Then, if $C$ is any abelian group and $f\colon A\to C$ and $g\colon B\to C$ are homomorphisms, then there is a unique homomorphism $h\colon A\oplus B\to C$ such that $$ h\circ i_A=f,\qquad h\circ i_B=g. $$

Once you have proved this property, then you have as a consequence that $$ \def\Hom{\operatorname{Hom}} \varphi\colon\Hom(A\oplus B,C)\to\Hom(A,C)\times\Hom(B,C) $$ defined by $$ \varphi(h)=(h\circ i_A,h\circ i_B) $$ is a bijection. Since $\Hom(\mathbb{Z}/p\mathbb{Z},\mathbb{Z}/p\mathbb{Z})$ has $p$ elements, you have the final answer.

This can be extended to the case of $\Hom(G\oplus G\oplus \dots\oplus G,H)$.


If you want to know how many homomorphisms are onto, the answer is quite easy, because $\mathbb{Z}/p\mathbb{Z}$ has just one proper subgroup, so a nonzero homomorphism is onto.


Can there be onto homomorphism $\mathbb{Z}\to S_3$? Of course not, because…

How many homomorphisms are there? The idea of considering the cyclic subgroups of $S_3$ is good.

Now, there are three subgroups of order $2$ and one of order $3$. Since $2$ and $3$ are prime, if $H$ is a subgroup with order $2$ or $3$, a nontrivial homomorphism $\mathbb{Z}\to H$ is surjective. So, for each of the three subgroups of order $2$ there is one nontrivial homomorphism; there are two for the subgroup of order $3$. So the number of homomorphisms is $$ 1+3+2=6 $$ ($1$ is for the trivial homomorphism).

Of course, the case of homomorphism from $\mathbb{Z}$ to $S_3$ is easy, because $\mathbb{Z}$ is free on one generator, so in general a homomorphisms from $\mathbb{Z}$ to $G$ is determined by selecting an element of $G$ to be the image of $1$. More complex reasoning would be needed in case the homomorphisms from $A$ (an abelian group) to $S_3$ are to be found.

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  • $\begingroup$ There can be no onto homomorphism $Z \rightarrow S_3$ because : let $\Psi : Z \rightarrow S_3$ be a homomorphism. Since, the kernel of a homomorphism is a normal subgroup of the domain, and every subgroup of $Z$ is of the form $nZ =>$ kernel of $Z$ must be of the form $nZ$ for some $n$. But $Z/nZ \approx Z_n$ and by the first isomorphism theorem : $Z/nZ$ should be isomorphic to $S_3$ but $S_3$ is not cyclic. Is this explanation correct? $\endgroup$ – MathMan May 31 '14 at 15:43
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    $\begingroup$ @VHP Easier: $S_3$ is not abelian. $\endgroup$ – egreg May 31 '14 at 15:46
  • $\begingroup$ So, for each of the three subgroups of order $2$ there is one nontrivial homomorphism; there are two for the subgroup of order $3$. I am afraid I don't understand this correctly enough. Can you give me a hint ? $\endgroup$ – MathMan May 31 '14 at 15:58
  • $\begingroup$ @VHP How many elements does $\operatorname{Hom}(\mathbb{Z},G)$ have, where $G$ is any finite abelian group? $\endgroup$ – egreg May 31 '14 at 16:00
  • $\begingroup$ Honestly, my book Gallian hasn't introduced a topic in this regard yet ( i.e. number of homomorphisms specifically for an abelian group ) but I can still try Suppose in $\Psi :G \rightarrow G'$ is a homomorphism, then if $a$ is a generator of $G$ then, number of homomorphisms is completely determined by the number of values $\Psi(a)$ can take which will be equal to the number of elements in $G'$ whose order divides the order of $a$ in $G$ In this case, there are 3 cyclic subgroups of order 2, 2 cyclic subgroups of order 3 and 1 trivial subgroup. hence, no of homomorphisms$= 3+2+1 = 6$ $\endgroup$ – MathMan May 31 '14 at 16:33
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Hint: To classify/count homomorphisms $G_1\to G_2$ between finite groups, it is often enough to look at the image of the generators of $G_1$, which will determine the whole homomorphism.

Examples:

  1. We want to find all homomorphism $\mathbb{Z}\to\mathbb{Z}_n$. The group $\mathbb{Z}$ is generated by the element $1$, and we have no relations (it is the free group on one element), thus $1$ can map to any element $x\in\mathbb{Z}_n$. This implies that we have exactly $n$ such homomorphisms, given by $k\mapsto nk$.
  2. Now we look for homomorphisms $D_3\to\mathbb{Z}_3$. The group $D_3=\left<r,s|r^3,s^2,(rs)^2\right>$ is finitely presented and generated by the two elements $r,s$. Notice that $r$ must map to an element of order $3$ (so anything goes), and $s$ to an element of order $2$ (so we must have $s\mapsto 0$). You can easily check that we have $3$ possible homomorphisms, given by $s\mapsto 0$ and $r\mapsto x$ with $x=0,1,2$.
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  • $\begingroup$ Since $Z_p \oplus Z_p$ is not cyclic, will the concept work in this question? $\endgroup$ – MathMan May 31 '14 at 14:24
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    $\begingroup$ yes, there are two generators, namely $(1,0)$ and $(0, 1)$. $\endgroup$ – Dustan Levenstein May 31 '14 at 14:25
  • $\begingroup$ Why did you go for $(1,0) , (0,1)$? $(1,1)$ also has order $p$? $\endgroup$ – MathMan May 31 '14 at 14:28
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    $\begingroup$ @VHP You can use either $\{(1,0),(0,1)\}$ or $\{(1,0),(1,1)\}$ or $\{(1,1),(0,1)\}$ as set of generators, the final result will not change. $\endgroup$ – Daniel Robert-Nicoud May 31 '14 at 14:33
  • $\begingroup$ So, $Z_p \oplus Z_p = \langle (1,0) \rangle \oplus \langle (0,1) \rangle $. $\Psi(1,0)$ and $\Psi(0,1)$ each should have order $p$. Hence, total homomorphisms possible number $p.p =p^2?$ For the case of homomorphisms $ \Psi :Z \rightarrow S_3$ , the generator $1$ can get mapped to either elements of orders $2$ or $3$. Is this correct? $\endgroup$ – MathMan May 31 '14 at 14:47
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Hints: I would be inclined to see what the kernel could be in the first question. It can't be the trivial group as the image group is not big enough. There is only one way the kernel can be the whole group. Otherwise, the kernel must be cyclic of order $p.$ How many subgroups of order $p$ are there, and how many different homomorphisms can have the same kernel of order $p$? For the second problem, the homomorphism is completely determined by the image of $1$. How many choices are there for the image of $1?$ ( I did not read teh question carefully enough- I did not notice that the homomorphism should be onto- neverthless, the answer as written makes sense: how many choices for the image of $1$ can make the homomorphism ONTO? The answer to that is clear).

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  • $\begingroup$ for the second problem, $\Psi(1)$ can take $6$ values since $|S_3|=6$ . Hence, there are 6 possible homomorphisms ? But, let $\Psi : Z \rightarrow S_3$ be a homomorphism. Since, the kernel of a homomorphism is a normal subgroup of the domain, and every subgroup of $Z$ is of the form $nZ =>$ kernel of $Z$ must be of the form $nZ$ for some $n$. But $Z/nZ \approx Z_n$ and by the first isomorphism theorem : $Z/nZ$ should be isomorphic to $S_3$ but $S_3$ is not cyclic. So, how can this homomorphism hold true? $\endgroup$ – MathMan May 31 '14 at 15:29
  • $\begingroup$ I have added something to my answer. $\endgroup$ – Geoff Robinson May 31 '14 at 15:35
  • $\begingroup$ So, the answer should be that there is only the trivial ontop homomorphism from $Z \mapsto S_3?$ because $Z/Ker~\Psi \approx S_3$ in only that case? where $Ker~\Psi$ is of the form $nZ$ $\endgroup$ – MathMan May 31 '14 at 15:39
  • $\begingroup$ Thank you for your answer. I guess I get the first 2 parts :) $\endgroup$ – MathMan May 31 '14 at 16:55

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