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How can we show that an abelian group of order less than $1024 $ has a set of generators of cardinality less than $10 $ ?

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Use the classification theorem. –  Chris Eagle Feb 29 '12 at 19:28
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Hint : $1024 = 2^{10}$. –  Joel Cohen Feb 29 '12 at 19:36
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4 Answers

up vote 6 down vote accepted

The Main theorem on the structure of finitely generated abelian groups tells you

that your group G is of the following form $$ G\cong C_{n_1}\times\ldots\times C_{n_k}, $$ where the $C_{n_i}$ are cyclic groups of order $n_i$ and the order of $G$ is therefore $\prod{n_i}<1024$. In this form the generators are easily determined (Hint: there are $k$ of them), so the larger $k$ the larger the number of generators. So we see that if we want to maximise the number of generators we should pick the $n_i$ as small as possible, $n_i=2$. Now $\#G=\prod{n_i}=2^k<2^{10}=1024$, so the maximum for $k$ is 9.

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I would have commented on Steven Stadnicki's post if I had enough reputation.

For $p$-groups $G$ there is the so called Burnside basis theorem (see http://planetmath.org/encyclopedia/BurnsideBasisTheorem.html), which asserts that a minimal generating set of $G$ has cardinality $\mathrm{log}_p(|G/\Phi(G)|)$, where $\Phi(G)$ denotes the Frattini-subgroup (i.e. the intersection of all maximal subgroups). It is a fact that $G/\Phi(G)$ is elementary abelian or equivalently a $\mathbb{F}_p$ vector space.

By this theorem you can see that a group of order 1024 has a minimal generating set containing 10 elements if and only if the group is elementary abelian. In all other cases a minimal generating set contains less elements.

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In fact, this can be shown straightforwardly even for non-abelian groups. Consider a minimal generating set $g_1, g_2, \ldots, g_n$ for any finite group $G$ (minimal here means that no generator in the set can be written in terms of the others) and consider the set of products $\displaystyle{\Pi_{i\in S}g_i}$ where $S$ ranges over the $2^n$ subsets of $\{1..n\}$ (and each product is taken in increasing order of $i$); e.g., $e$ (the empty product), $g_1,\ g_2,\ g_1g_2,\ g_3,\ g_1g_3,\ \ldots$ Can you show that each of these $2^n$ elements of $G$ is distinct? (Hint - suppose two elements are equal and consider the first generator that's in one but not the other.) And then invert this to find a statement about the cardinality of a minimal generator set for the group?

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+1 for simplicity, generality, and generation from first principles. –  John Bentin Feb 29 '12 at 20:31
    
Indeed a much better answer than mine. –  Michalis Mar 5 '12 at 10:11
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One knows that any finite abelian group $G$ is product of cyclic groups: $G=C_1\times\cdot\times C_t$. If $g_i$ is a generator of $C_i$ then the set of all the $g_i$ generates $G$.

Now, since $|C_i|\geq2$ and $2^{10}=1024$, we must have $t<10$.


Michalis arrived first......

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