Direct limit of a system: computation

Question

I am considering a homomorphism from $\mathbb{Z}_4\rightarrow \mathbb{Z}_6$ given by $\bar{1}\mapsto \bar{3}$. My question is:

What is the direct limit of this system in the category of abelian groups and in the category of all groups.

In abelian groups, it is the quotient $\mathbb{Z}_4\oplus \mathbb{Z}_6/N$ where $N=\{(0,0), (1,3), (2,0), (3,3)\}$. I think then $\mathbb{Z}_4\oplus \mathbb{Z}_6/N$ is isomorphic to $\mathbb{Z}_6$, is it correct?

In the category of all groups, I do not understand the direct limit of above system. Can one help for it? Hint?

Answer 1

Yes, the direct limit of $f: \mathbb{Z}/4 \to \mathbb{Z}/6$ is $\mathbb{Z}/6$. You can take the (surjective) map $\mathbb{Z}/4 \oplus \mathbb{Z}/6 \to \mathbb{Z}/6$ given by $(a, b) \mapsto f(a) + b$, and $N = \{(a, -f(a)\}$ would exactly be the kernel.

In any category, it is true that the direct limit of $A \to B$ is $B$, but in the category of (not necessarily abelian) groups, it is no longer true that $\lim_{\to}\limits (A \to B) = (A \oplus B)/N$, unless by $A \oplus B$ you mean the categorical direct sum (ie the free product, not the direct product) and replace $N$ by the normal subgroup generated by $\{a f(a)^{-1}\}$.

Answer 2

$$ G=<x,y:x^4,y^6,xy{3}> $$ I may know how to explain to you. Every word of $G$ must be ${x}^{a_1}y^{b_1}\cdots x^{a_n}y^{b_n} $ . When $x$ is replaced by $y^3$ , you find that $G$ is abelian ,even cyclic and every word is $y^{m}$ for some $n=0,1,2,3,4,5$. Hence, your result gets back the abelian case. So still $G=\mathbb{Z}_6$. Hope it helps you.