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I wonder if the statement in the title is true, so if:

$$G=\langle g_1,g_2,\ldots,g_n\rangle$$ means that we can write every subgroup $H$ of $G$ as $$H=\langle h_1,h_2,\ldots,h_m\rangle$$ for some $h_i\in G$ and $m\leq n$.

What I'm thinking so far: if $n=1$, then: $$H=\langle h_1,h_2\rangle=\langle g^a,g^b\rangle=\langle g^{\gcd(a,b)}\rangle$$ So the statement is true for $n=1$.

However for $n>1$ I'm having trouble proving/disproving it. I think it comes down to some number theory stuff that I'm not aware of. For example for the case $n=2$:

Say $|g_1|=n, |g_2|=m$

$$H=\langle h_1,h_2,h_3\rangle =\langle g_1^{a_1} g_2^{a_2}, g_1^{b_1} g_2^{b_2}, g_1^{c_1} g_2^{c_2}\rangle$$ So I think that proving that this can be generated by only $2$ of the elements comes down to proving that there are $x,y$ s.t.

$$xa_1+yb_1\equiv c_1\bmod n$$

$$xa_2+yb_2\equiv c_2\bmod m$$ However, I don't know the conditions that have to be satisfied for this system to have solutions.

By the way, I'm not sure/given that this is actually true.

edit: by the way, I think that in writing down that system I accidentally assumed the group to be Abelian. Now I'm interested in the result both for Abelian and non-Abelian groups

user2520938
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  • That sounds like an answer to me. – theage Mar 09 '15 at 21:16
  • @Diego If you can give me hint (just a small one preferably) on how to prove this for Abelian groups, and write it down in an answer, that would be nice. – user2520938 Mar 09 '15 at 21:18
  • I changed $G=<g_1,g_2,...,g_n>$ to $G=\langle g_1,g_2,\ldots,g_n\rangle$. Not only the angle brackets, but also the dots. If you write g_2,...,g_n instead of g_2,\ldots,g_n in actual LaTeX rather than MathJax (which is what is used here), then you'll see $g_2,\text{...},g_n$ instead of $g_2,\ldots,g_n$. ${}\qquad{}$ – Michael Hardy Mar 09 '15 at 21:21

3 Answers3

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The statement is true for abelian groups: every abelian group generated by $n$ elements is a quotient of $\mathbb{Z}^n$ and every subgroup of $\mathbb{Z}^n$ is free generated by at most $n$ elements. This can be found in any text elementary group theory.

It is false for non abelian groups. For example the permutation group $S_n$ is generated by two elements: the cycle $(12\ldots n)$ and the trasposition $(12)$. Every finite group $G$ is a subgroup of some $S_n$ but not every finite group is generated by two elements

Diego
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3

Take the free group generated by two symbols $a,b$, i.e. the group of all strings that are made from $a,b,a^{-1},b^{-1}$ symbols where you cancel out substrings like $aa^{-1}$ when you express or multiply. This is generated by $a,b$ (assuming that inverses are allowed when forming the group). However you can find subgroups that require arbitrarily many generators or even infinitely many to generate. For example, the subgroup generated by $a^nb$ for all $n \geq 1$ requires infinitely many generators.

user2566092
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This is not true for finite (non-abelian) simple groups and their maximal subgroups.

It is known that every finite (non-abelian) simple group can be generated by $2$ elements. (This was proved by several people, e.g. Miller, Steinberg, Aschbacher and Guralnick.)

In the paper Generation and random generation: from simple groups to maximal subgroups by Burness, Liebeck and Shalev the authors prove the the following:

Theorem. Every maximal subgroup of a finite simple group is $4$-generator.

This means that every maximal subgroup is generated by at most $4$ elements. To answer your question, the authors comment that there are infinitely many examples in which a maximal subgroup needs at least $4$ elements to be generated. See the paper for the explicit examples.

rafforaffo
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