## Hi! Some problems in algebra.

My first time posting here! Wish all of you a merry Christmas! I am alone in my apartment, reviewing the homework for the semester and found some problems…

So we have an interesting fact: a subgroup of a finitely generated group is not necessarily finitely generated (a nice exercise by considering the commutator subgroup of the free group of two generators).

So my questions are:

1. Show that if a finitely generated abelian group $G$ is generated by $n$ elements, then any subgroup of $G$ is generated by $\leq n$ elements. (This is a mid-term problem, and I forgot how to do it…)
2. Is the same statement true for finite but non-abelian group? (Open to me.)

Thanks!

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### 4 Responses to Hi! Some problems in algebra.

1. anthonmak says:

Oh I got (1). Sorry about that!

One can show that there exists a surjective homomorphism $\phi: \mathbb{Z}^n \to G$, either directly or from the universal property of free abelian groups. Then, if $H$ is any subgroup of $G$, then $\phi^{-1}(H)$ is a subgroup of $\mathbb{Z}^n$. Then we use the fact that any subgroup of $\mathbb{Z}^n$ must be isomorphic to $\mathbb{Z}^r$ where $r \leq n$. Then $H$ must be generated by $\leq r$ elements.

This argument does not apply to (2) because a surjective homomorphism from a non-abelian group to a subgroup needs not exist, say $S_3$ to $\mathbb{Z}_3$, by order consideration.

2. kingleunglee says:

wait, order of $S_3=6$, and $\mathbb{Z}_3$ is a normal subgroup of $S_3$

3. KKK says:

If there’s such surjective homomorphism $\phi$, we will draw a contradiction as follows.
$\mathbb{Z}_3$ has exactly one order 1 element and two order 3 element. For each transposition $\sigma$ of $S_3$ (i.e. $\sigma$ is of the form $(i, j), i\neq j\in \{1, 2, 3\}$), as $ord(\sigma)=2$, the order of $\phi(\sigma)$ must divide 2, so it must be 1 (since 3 does not divide 2). So we conclude that $\phi(\sigma)=id$. As $S_3$ is generated by all the transpositions, $\phi$ must be trivial, which contradicts its surjectivity.

4. KKK says:

The answer to (2) is no in general.
Take $G=S_6$, the permutation group of six elements, then it can be generated by two elements: $(1\, 2)$ and $(1\, 2\, 3\, 4\, 5\, 6)$, this is quite clear because each transposition $\sigma$ of the form $\sigma=(i, i+1), 1\leq i\leq 5$ can be generated by these two elements.
On the other hand, consider the subgroup $H$ generated by $\sigma_1=(1\, 2), \sigma_2=(3 \,4)\text{ and }\sigma_3=(5\, 6)$. We clearly have $\sigma_i^2=id$ and $\sigma_i\sigma_j=\sigma_j\sigma_i$, from this it is easy to see that $H\cong \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$, whose number of generators is at least 3.