Let N be a cyclic normal subgroup of a finite group G and $\frac{G}{N}$ be a p-group in which p and the order of N is coprime. Then what we can say about the relation between the minimum number of generators of G and $\frac{G}{N}$?
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1$\begingroup$ I believe Schur-Zassenhaus implies the number of generators of $G$ is at most the sum of the number of generators of $N$ + the number of generators of $G/N$. On the other hand, any set of generators of $G$ induces a set of generators of $G/N$. So the two numbers differ by at most $1$. $\endgroup$– WojowuCommented Oct 21, 2016 at 11:40
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$\begingroup$ what we can say about the exact minimum number of generators? $\endgroup$– Mojtaba JazaeriCommented Oct 21, 2016 at 11:43
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5$\begingroup$ The answer to your question is that they differ by $0$ or $1$, and both are possible. You could say more under certain conditions. For example, if $G/N$ is not cyclic and $|N|$ is divisible by at most two primes, then the generator numbers are equal. But you need to ask a more precise question if you want a more detailed answer. $\endgroup$– Derek HoltCommented Oct 21, 2016 at 13:40
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2$\begingroup$ I think you could state it this way, not that it differs much from what is said already. Let $d(X)$ denote the minimum number of generators of a finite group $X$. Then $d(G) = 1 + d(G/N)$ if and only if some generating set of cardinality $d(G)$ of $G$ contains an element of $N.$ Otherwise $d(G) = d(G/N).$ $\endgroup$– Geoff RobinsonCommented Oct 23, 2016 at 12:54
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1 Answer
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Let $d(G)$ be the minimum number of generators for a group $G$.
Now, $d(G) \leq d(G/N) + 1$. Let $K$ be a set of representative elements corresponding to a minimum generating set for $G/N$; so, $|K| = d(G/N)$. Now, given $g \in G$, there exists a $k$, which is a product of elements in $K$, such that $g \in kN$; so, we need only take the generator of $N$ to the right power to construct $g$.
Further, $d(G/N) \leq d(G)$ because, given a generating set for $G$, we need only take its image under the quotient map to get a generating set for $G/N$.
Together, we get that $|d(G/N) - d(G)| \leq 1$. Note that this holds for every normal subgroup $N$! In fact, there is the general result that if $H \subset G$ is a subgroup, the $d(G) \leq (G \colon H) + d(H) - 1$, which is a bit weaker than our result above because we can't take $d(G/H)$ for an arbitrary subgroup.
