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Show that if $G$ is a finitely generated abelian group, every subgroup of $G$ is finitely generated.

Proof: Suppose that $G$ is a finitely generated abelian group. This means that for each $g\in G$ we may write a finite sum $g=\sum n_{\alpha}g_{\alpha}$ and the set $\{g_{\alpha}\}$ that generates $G$ is finite. Let $T \subseteq G$ be a subgroup of $G$. If $T$ is the trivial subgroup then it is finitely generated. Suppose $g\in T$. This means $g\in G$. Since $G$ is a finitely generated abelian group we know that we may write a finite sum $g=\sum n_{\alpha}g_{\alpha}$ and the set $\{g_{\alpha}\}$ generates $G$ is finite, which means we may do the same for $T$. Therefore, $T$ is finitely generated and we have shown every subgroup of $G$ is finitely generated. $\therefore$

I am a little rusty writing proofs, even rather simple ones of this kind. I was hoping for some proof verification regarding this straightforward exercise. Thank you for your time.

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  • $\begingroup$ It's actually a bit more subtle than this. You've not exhibited a single set that generates $T$: you could be using vastly different generating elements of $G$ for different $g \in T$. And if you start throwing them all in haphazardly, you might end up generating something bigger than $T$. $\endgroup$
    – Randall
    Nov 28, 2017 at 15:53
  • $\begingroup$ Also, if you don't find yourself making use of the abelian assumption, something is wrong because the corresponding statement for generic groups is really false. $\endgroup$
    – Randall
    Nov 28, 2017 at 15:54
  • $\begingroup$ @Randall You're right, I'm trying a different route. Since we know all subgroups of Abelian groups are normal we have that, for a normal subgroup $T$ of $G$ with cosets $x,y$, $(xT)(yT)=(xy)T$. The set of cosets form a quotient group, $G/T$. The homomorphism $f:G\rightarrow G/T$ given by $f(a)=aT$ sends subgroups of $G$ to subgroups of $G/T$ $\endgroup$ Nov 28, 2017 at 16:20
  • $\begingroup$ The kernel is always normal and the image $f(G)$ is isomorphic to $G/ker(f)$ by the first isomorphism theorem, the correspondence is a bijection between the set of all quotient groups of $G/N$ and the set of homomorphic images of $G$. Then the kernel of the quotient map is $T$ itself, so the normal subgroups are the kernels of homomorphisms with domain $G$, which means that these generating elements of $G$ will be preserved in every subgroup of $G$ since those subgroups must be normal. $\endgroup$ Nov 28, 2017 at 16:24

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Proof: Suppose that $G$ is a finitely generated abelian group. This means that every subgroup of $G$ is normal. If a subgroup is a trivial subgroup we know that it is finitely generated. Similarly, if a subgroup is an improper subgroup we also know that it is finitely generated. For a normal subgroup $N$ of $G$, we may define multiplication over their cosets $a_{1}$ and $a_{2}$ in the following manner: \begin{equation} (a_{1}N)(a_{2}N)=(a_{1}a_{2})N \end{equation} This turns the set of cosets into a group called the quotient group, $G/N$. We define the natural homomorphism $f: G\rightarrow G/N$ given by $f(g)=gN$ for each $g\in G$. By the first isomorphism theorem we have that this image $f(G)$ is isomorphic to $G/\ker(f)$. The kernel of $f$ is $N$. Then we have that $f(G)$ is isomorphic to $G/N$. This correspondence is a bijection between the set of all quotient groups $G/N$ and the set of all homomorphic images of $G$ up to isomorphism. This means that normal subgroups are the kernels of homomorphisms with domain $G$.

Now we will show that the kernel of every group homomorphism is finitely generated. Let $e_{G/N}$ be the identity of the quotient group $G/N$ and let $\ker{f}=\{g\in G: f(g)=e_{G/N}\}$. Since $G$ is finitely generated we know that the quotient group $G/N$ is also finitely generated. Also, since $G$ is finitely generated we know that for any element $g \in G$ we may write $g=a_{1}x_{1}+\cdots+ a_{n}x_{n}$ for integers $a_1\dots a_{n}$ and finite generators $x_{1}\dots x_{n}\in G$. Then this means we may do the same for our quotient group $G/N$, which means we may write our identity $e_{G/N}=b_{1}y_{1}+\cdots+ b_{n}y_{n}$ for integers $b_1\dots b_{n}$ and finite generators $y_{1}\dots y_{n}\in G/N$. We already confirmed that the kernel of $f$ is $N$, which means that $N=\{ g\in G:f(g)=e_{G/N}\}$. Since the identity of $G/N$ can be written as a finite sum of generators with weights we have that $N$ is finitely generated. Thus we have shown that every subgroup of $G$ is finitely generated. $\therefore$

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  • $\begingroup$ Now to to show that the kernel of every group homomorphism is finite generated... $\endgroup$ Dec 13, 2017 at 2:42
  • $\begingroup$ @AndresMejia I tried to complete the proof, when you get the chance can you let me know if there needs more improvement? Thanks man $\endgroup$ Dec 13, 2017 at 16:31

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