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I have been self-studying from Aluffi's Algebra: Chapter 0. I am looking at Chapter II section 3, exercise 3.8. Here it is:

As far as I can tell, I need to do the following:

  1. Describe the projections $\pi_2 : C_2 \to G$ and $\pi_3 : C_3 \to G$.
  2. Show that the universal property for coproducts is satisfied: that is, for any group $A$ and any choice of group homomorphisms $\varphi_2 : C_2 \to A$ and $\varphi_3 : C_3 \to A$, there exists a unique group homomorphism $\sigma : G \to A$.

Giving a more concrete description of the cyclic groups, take $C_2 = \{0, 1\}$ and $C_3 = \{0, 1, 2\}$. Then the projections are $\pi_2(k) = x^k$ where $k=0,1$, and $\pi_3(l) = y^l$ where $l=0,1,2$. I think that this takes care of (1).

Beyond this, it is not clear to me what I should do. Any help would be appreciated!

JJH
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3 Answers3

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You can look at $G$ as a group that has strings like $xyxy^2xy$ or $y^2xyxy$ as elements.

Then $\phi:G\to A$ on e.g. $xyxy^2xy$ must be prescribed by:$$xyxy^2xy\mapsto\phi_2(x)\phi_3(y)\phi_2(x)\phi_3(y)^2\phi_2(x)\phi_3(y)$$

This $\phi$ must be shown to be a group homomorphism with $\phi\circ\pi_2=\phi_2$ and $\phi\circ\pi_3=\phi_3$, and must be shown to be unique in satisfying this.

drhab
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  • Sorry, but using OP's notation, the image of $xyxy^2xy$ should be like: $\phi_2(1)\phi_3(1)\cdots$. I just post an answer clarifying yours. Is that correct? – Hamilton Feb 02 '21 at 10:00
  • @YarenZhang Indeed I identified the groups $C_2$ and $C_3$ with the subgroups $$ and $$. Your answer looks okay to me. – drhab Feb 02 '21 at 11:27
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A map out of a group is just given by a map on the generators such that the relations are satisfied. Thus a map from $G$ to $H$ is uniquely determined by two elements of $H$, one which squares and one which cubes to the identity. This is exactly the description of a pair of maps from $C_2$ and $C_3$ to $H$, and the correspondence is induced by maps from $C_2$ and $C_3$ to $H$ taking a generator to a generator, as you say.

Kevin Carlson
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I don't know if you are satisfied with the above two brilliant and suggestive answers given by @Kevin Arlin and @drhab. So I will elaborate on @drhab's answer, using your notations.

You can look at $G$ as a group that has strings like $xyxy^2xy$, $y^2xyxy$, or $xyxy^2$ as elements. Take $xyxy^2$ as an example.

  1. The condition "$\sigma:G\to A$ is a group homomorphism" forces that $$xyxy^2 \mapsto\sigma(x)\sigma(y)\sigma(x)\sigma(y)^2$$

  2. Besides, the condition "the diagram commutes such that $\sigma \circ \pi_2 = \varphi_2$ and $\sigma \circ \pi_3 = \varphi_3$" and the fact that $\pi_2(1)=x$ and $\pi_3(1)=y$ forces the above map becomes

$$xyxy^2 \mapsto\sigma(x)\sigma(y)\sigma(x)\sigma(y)^2 = \sigma \circ \pi_2(1)\cdot \sigma \circ \pi_3(1) \cdot \sigma \circ \pi_2(1) \cdot (\sigma \circ \pi_3(1))^2 = \varphi_2(1)\varphi_3(1)\varphi_2(1)\varphi_3(1)^2$$

In fact, the map $\sigma: G\to A$ is completely prescribed by the above 2 "forces". So the map $\sigma: G\to A$ is the unique group homomorphism.

The point behind the proof is just what @Kevin Arlin said in his answer: the group homomorphism $\sigma: G\to$ any set $A$ is completely determined by how the homomorphism act on the group generators (Meanwhile, it must satisfy all relations. See this post). Your functions $\pi_2$ and $\pi_3$ are "injections", they identify $[1]$ in $C_2$ with $x$ in $G$, $[1]$ in $C_3$ with $y$ in $G$. After this identification, $\varphi_2$ and $\varphi_3$ actually specify how $\sigma$ should act on generators $\{x,y\}$ in $G$. So $\sigma$ has no choice, as has been proved above.

Kevin
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Hamilton
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  • I am also studying Aluffi's book and wanted to clarify one thing. Are we assuming that two different strings are always distinct? Otherwise, it is necessary to show that $\sigma$ is well-defined. For example, if $xy = y^2x$, then we have to show that $\sigma(x)\sigma(y) = \sigma(y)^2\sigma(x)$. – Yerbolat Jul 21 '21 at 15:35