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If $A$ is a f.g. abelian group, let $\mu(A)$ be the minimal number of elements needed to generate $A$.

If we know $\mu(A)$, what is $M(A)=\sup\limits_{B\subseteq A}\mu(B)$, with the supremum taken over all subgroups $B\subseteq A$?

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I think you mean, "proper subgroups." Otherwise, it's clear that $M(A)=\mu(A)$. – goblin Feb 17 '13 at 9:50
@user18921, using exactly the same definitions for non-abelian groups, it is easy to see that $\mu(S_{2n})=2<n\leq M(S_{2n})$ as soon as $n>2$, by considering the subgroup of $S_{2n}$ generated by the $n$ transpositions of the form $(i,i+n)$ with $i\in\{1,\dots,n\}$. Any clear reason would have to become less clear in the non-abelian case. – Mariano Suárez-Alvarez Feb 18 '13 at 5:34
up vote 4 down vote accepted

I think $M(A) = \mu(A)$.

Fix a subgroup $B \leq A$, and write $A$ as a the direct sum of cyclic groups. We try to show that $\mu(B) \leq \mu(A)$. Let $a_1,\cdots,a_n$ be a generator of those cyclic groups. We know that $B$ is finitely generated (since $A$ is a Noetherian $\mathbb{Z}$-module). Pick one such generating set $b_1,\cdots,b_m$, and write $b_i = \sum c_{ij} a_j$ for each $i$, where $c_{ij} \in \mathbb{Z}$.

Now Smith normal form says that the matrix $(c_{ij}) = PDQ$, where $D = diag(d_1,\cdots,d_r,0,\cdots,0)$ is diagonal ($r \leq \min(m,n)$), and $P,Q$ are invertible in $GL_n(\mathbb{Z})$.

Let $Q = (q_{kl})$, and consider another generating set of $A$ $a_1',\cdots,a_n'$ defined by $a_i' = \sum q_{ij} a_j$. Then $B$ is generated by $d_1a_1',\cdots, d_r a_r'$, where $r \leq \min(m,n) \leq n$, which shows that $\mu(B) \leq \mu(A)$.

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Consider $\mathbb Z_p \times \mathbb Z_p$ every subgroup has order $p$ and is generated by a single element. But the group is generated by two elements. – JSchlather Feb 17 '13 at 7:16
@JacobSchlather, the whole group is trivially a subgroup as well. – user27126 Feb 17 '13 at 7:21
I see, I misread it as being proper subgroups. – JSchlather Feb 17 '13 at 7:22

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