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.)


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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.

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