Minimum size of generating set of subgroup may be more than of whole group

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Statement

It is possible to have a finite group $G$ and a subgroup $H$ with the property that the Minimum size of generating set (?) of $G$ is less than that of $H$.

Related facts

Opposite facts

• Cyclicity is subgroup-closed: In particular, this means that any group where the minimum size of generating set is bigger than 1 cannot be put in a group where the minimum size of generating set is 1 or less.

Facts used

1. Every finite group is a subgroup of a finite 2-generated group

Proof

Counterexample with smallest order

Further information: SmallGroup(16,3), subgroup structure of SmallGroup(16,3)

This is the smallest order counterexample both for $G$ and for $H$.

Let $G$ be SmallGroup(16,3), i.e., it is a group of order 16 given by the presentation:

$G := \langle a,b,c \mid a^4 = b^2 = c^2 = e, ab = ba, bc = cb, cac^{-1} = ab \rangle$

Note that although the above presentation uses three generators, the minimum size of generating set for $G$ is 2, because it is generated by $a$ and $c$ alone, because $b = a^{-1}cac^{-1}$ by the final relation in the presentation.

Consider the subgroup:

$H := \langle a^2,b,c \rangle$

It is easy to check that $H$ is isomorphic to elementary abelian group:E8, and in particular its minimum size of generating set is 3, which is bigger than that of 2.

Explanation for why no example of smaller order is possible: Because cyclicity is subgroup-closed, it is not possible to get an example with $G$ cyclic. Thus, $G$ must have a minimum size of generating set of size at least 2. To yield a counterexample, $H$ must have a minimum size of generating set at least 3. We know that the minimum size of generating set is bounded from above by the sum of exponents of the prime divisors in the prime factorization of the order, so the smallest possible order allowing a minimum size of generating set $3$ is $2^3 = 8$. The only group of this order satisfying the necessary condition is elementary abelian group:E8. Since the order of $H$ divides the order of $G$, but they cannot be equal, the order of $G$ is at least $16$. Of the groups of order 16 that contain this as a subgroup, SmallGroup(16,3) is the only one having a generating set of size two.

Construction of general family of counterexamples

Fact (1), along with the observation that we can construct finite groups with arbitrarily large minimum size of generating sets (e.g., elementary abelian groups of arbitrarily large rank), gives us a large and infinite colleciton of examples.

Fact (1) is typically proved using symmetric group on a finite set as the large group in which the embedding is done. The smallest example of such an embedding is the embedding of elementary abelian group:E8 in symmetric group:S6 (this is not the Cayley embedding; the Cayley embedding would be in symmetric group:S8).