Categorical notes on groups

Products and Coproducts

$\text{Gp}$ has products, and the product of two groups $G,H$ is supported on the product $G\times H$ of the underlying sets. Componentwise multiplication defines a group structure on $G\times H$, and it is easy to verify that this group is actually a product in $\text{Gp}$ ( Aluffi (2009) II.3.4).

$\text{Gp}$ also has coproducts: See Aluffi (2009) Exercises 5.6, 5.7, 8.7.

The nice thing about $\text{Ab}$ is that finite coproducts in $\text{Ab}$ coincide with finite products.

Proposition 31.1(Proposition).

If $G$ and $H$ are abelian groups, then the product $G\times H$ satisfies the universal property for coproducts in $\text{Ab}$.

Proof.

We need to show that$(G\times H,i_G,i_H)$ is initial in ${\text{C}}^{G,H}$. Let $(A,{\phi }_G,{\phi }_H)\in {\text{C}}^{G,H}$. We need a unique homomorphism $\phi$ which makes the following diagram commute:

We are forced to define $\phi (g,e_H)\equiv {\phi }_G(g)$ and $\phi (e_G,h)\equiv {\phi }_H(h)$ for all $g\in G$ and $h\in H$. Since we require $\phi (g,h)=\phi (g,e_H)\phi (e_G,h)$ (and every element of $G\times H$ can be expressed uniquely as such a product), our definitions determine $\phi$ over its entire domain. So, we have exactly one candidate for $\phi$. It it a homomorphism?

$$\begin{array}{rl}\phi (a,b)\phi (c,d)& ={\phi }_G(a){\phi }_H(b){\phi }_G(c){\phi }_H(d)\\ & \stackrel{!}{=}{\phi }_G(a){\phi }_G(c){\phi }_H(b){\phi }_H(d)\\ & ={\phi }_G(ac){\phi }_H(bd)\\ & =\phi (ac,bd)\\ & =\phi ((a,b)(c,d)).\end{array}$$

The marked equality is true since $A$ is abelian.□

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When working as a coproduct, the product $G\times H$ of two abelian groups is often called their direct sum and is denoted $G\oplus H$.

Remark 31.2(Remark).

Products and coproducts are not the same when the index set is infinite.

1. The product ${\prod }_{i\in I}{A}_i$ is the group of all tuples $(a_i{)}_{i\in I}$ with componentwise addition.
2. The coproduct ${⨁}_{i\in I}{A}_i$ is the subgroup of the product consisting of tuples with finite support.

It is easily seen that the prescription in $(1)$ satisfies the universal property of products in $\text{Ab}$.

To prove $(2)$, let $Q$ be the subgroup of $P={⨁}_{i\in I}{A}_i$ generated by the images of the $A_i$. Then the universal property of $P$ induces a canonical homomorphism $\phi :P\to Q$. By extending the diagram to the right by the inclusion $\iota :Q\hookrightarrow P$, one observes that $\iota \circ \phi$ must be the identity. This forces $Q=P$.

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Homomorphisms

Proposition 31.3(Aluffi (2009) II.6.6).

Let $\phi :G\to G^{\prime }$ be a homomorphism. Then the inclusion $i:\ker \phi \to G$ is final in the category of group homomorphisms $\alpha :K\to G$ such that $\phi \circ \alpha$ is the trivial map.

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In other words, every group homomorphism $\alpha :K\to G$ such that $\phi \circ \alpha$ is the trivial homomorphism factors uniquely though $\ker \phi$. Note that $\overline{\alpha }:K\to \ker \phi$ is just $\alpha$ with restricted target.

Monomorphisms and epimorphisms

Proposition 31.4(Aluffi (2009) II.6.12).

Let $\phi :G\to G^{\prime }$ be a homomorphism. The following are equivalent:

1. $\phi$ is a monomorphism;

2. $\ker \phi =\{e_G\}$;

3. $\phi$ is injective (as a set-function).

Proof.

$(1)⟹(2)$: Consider the parallel compositions

$$\ker \phi \underset{e}{\stackrel{i}{\rightrightarrows }}G\stackrel{\phi }{\to }{G}^{\prime },$$

where $i$ is in the inclusion and $e$ is the trivial map. Both $\phi \circ i$ and $\phi \circ e$ are the trivial map; since $\phi$ is a monomorphism, this implies $i=e$, which implies $(2)$.□

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The analog of the above statement holds true for epimorphisms: a homomorphism $\phi :G\to G^{\prime }$ is an epimorphism iff it is surjective. However, Aluffi claims proving epimorphism $⟹$ surjective in $\text{Gp}$ is cumbersome, and ==only provides a proof in $\text{Ab}$==. He does so by defining cokernels in $\text{Ab}$, the universal property for which is obtained by reversing the arrows in the universal property of kernels:

Definition 31.5(Definition).

Let $\phi :G\to G^{\prime }$ be a group homomorphism. $\text{coker}\phi$ is the group equipped with a homomorphism $\pi :G^{\prime }\to \text{coker}\phi$ which is initial with respect to all morphisms $\alpha$ such that $\alpha \circ \phi =0$.

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Cokernels exist in $\text{Ab}$ because $\mathrm{Im}\phi ◃G^{\prime }$. The condition that $\alpha \circ {\phi }^{\prime }$ is trivial says that $\mathrm{Im}\phi \subseteq \ker \alpha$, and hence

$$\frac{G^{\prime }}{\mathrm{Im}\phi }\cong \text{coker}\phi$$

satisfies the universal property ($\overline{\alpha }$ is unique because of the universal property of quotient groups).

We can now state the analog of Prp 4:

Proposition 31.6(Aluffi (2009) II.8.18).

Let $\phi :G\to G^{\prime }$ be a homomorphism of abelian groups. The following are equivalent:

1. $\phi$ is an epimorphism;

2. $\text{coker}\phi$ is trivial;

3. $\phi :G\to G^{\prime }$ is surjective (as a set function).

Proof.

$(1)⟹(2)$: Consider the parallel compositions

$$G\stackrel{\phi }{\to }{G}^{\prime }\underset{\pi }{\stackrel{e}{\rightrightarrows }}\text{coker}\phi ,$$

where $\pi$ is the canonical projection. Both $\pi \circ \phi$ and $e\circ \phi$ are trivial, which implies $\pi =e$, which implies $(2)$.□

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Remark 31.7(Remark).

The problem in $\text{Gp}$ is that $\mathrm{Im}\phi$ is not guaranteed to be normal in $G^{\prime }$. However, cokernels may still be defined in $\text{Gp}$: the universal property is satisfied by $G^{\prime }/N$, where $N$ is the smallest normal subgroup of $G^{\prime }$ containing $\mathrm{Im}\phi$. But Prp 6 fails, because the implication $(2)⟹(3)$ fails.

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References

Aluffi, P. (2009). Algebra: Chapter 0. American Mathematical Society.