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For infinite groups (ex. Free Groups), a subgroup may have more number of generators than the group. For finite groups, is the number of generators of subgroup less than number of generators of group?

[Of course, we understand number of generators to mean "minimum number of generators"].

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Here's a counterexample (of minimal order!) on Vipul Naik's wiki, which is always good for this sort of thing:… – Dylan Moreland Sep 7 '11 at 5:05
up vote 7 down vote accepted

Take any group that cannot be generated by less than $n$ generators, e.g. $\overbrace{\mathbb{Z}/2\mathbb{Z}\times\ldots \mathbb{Z}/2\mathbb{Z}}^n$. Embed it into a symmetric group, e.g. by letting it act on itself by permutations. As you know, any symmetric group $S_m$ can be generated by two elements, a 2-cycles and an $m$-cycle. This way you get arbitrarily "bad" counterexamples to your question.

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