Suppose $G$ is a finite abelian $2$-group, and $S$ is a subset of $G$, $\langle S\rangle=G$,$S^{-1}=S$,$e\notin S$. How to determine whether there exists a maximal subgroup $M$ of $G$, such that $S$ is contained in $G\setminus M$?
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5$\begingroup$ Write $G$ additively (and assume $G\neq\{0\}$, as the answer is no for $G=\{0\}$). Let $N$ be the subgroup generated by $2G\cup\{s-t:s,t\in S\}$ (so $N$ has index at most 2). Then $M$ exists iff $N\cap S=\emptyset$ (in which case the only possible $M$ is $N$ itself). $\endgroup$– YCorCommented May 7 at 6:51
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$\begingroup$ @YCor why N has index at most 2? $\endgroup$– lunch zhengCommented May 7 at 14:42
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$\begingroup$ Because modulo $N$, the set $\{s-t:s,t\in S\}$ is trivial, which means that modulo $N$, $S$ is a singleton. So $G/N$ is generated by a singleton, hence cyclic, since $2G\subset N$, we deduce that $G/N$ has order $1$ or $2$. $\endgroup$– YCorCommented May 7 at 17:54
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$\begingroup$ @YCor Sorry，I understand $2G$ is the Frattini subgroup of $G$, so $G/2G$ must be an elementary abelian $2$-group. However, I'm not quite sure about the role of the set $\{s-t: s, t \in S\}$. Why is the subgroup $N$ generated by $2 G \cup\{s-t: s, t \in S\}$, either $G$ or a maximal subgroup? $\endgroup$– lunch zhengCommented May 8 at 7:40
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1$\begingroup$ Note that my first comment (with $N$ defined as the "normal subgroup generated by all squares and $\{st^{-1}:s,t\in S\}$") works assuming that $G$ is an arbitrary finite group. A restatement is that there exists $N$ with the given condition iff $S$ is contained in the nontrivial coset of some subgroup of index 2. $\endgroup$– YCorCommented May 8 at 12:27
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2 Answers
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Here's an algorithm. Choose an element $s\in S$ and write down a list $A$ of the elements $st$ with $t\in S\setminus \{s\}$. Next, write down a basis $B$ for $\operatorname{\rm Hom}(G,\mathbb{Z}/2)$. This gives you a matrix whose columns are indexed by $A$ and rows by $B$, giving the values of the homomorphisms on the elements. Then the question amounts to whether the rows are linearly independent. So do a row reduction and see whether you get a zero row. Keeping track of the row reduction will give you a linear combination of your basis of homomorphisms, and the kernel is then your maximal subgroup $M$. This doesn't depend on $G$ being abelian.
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$\begingroup$ but for any $G$ , how can i know the basis $B$ for $\operatorname{Hom}(G, \mathbb{Z} / 2\mathbb{Z})$. $\endgroup$ Commented May 8 at 5:59
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$\begingroup$ Mod out squares and commutators, to get an elementary abelian $2$-group. Then it's a vector space over $\mathbb{F}_2$. How are you given your group? $\endgroup$ Commented May 8 at 6:01
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2$\begingroup$ (actually, every commutator in any group is a product of three squares, so you can just mod out squares) $\endgroup$ Commented May 9 at 7:42
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Another way to look at it (given all the hypotheses), is that there is such a maximal subgroup $M$ if and only if there is an element $t \in S$ such that $St = \{st: s \in S \}$ does not generate $G$.
For if there is such an element $t$, take a maximal subgroup $M$ containing $\langle St \rangle.$ Then $t \not \in M$, for otherwise $t^{-1} \in M$ and $S = (St)t^{-1} \subseteq M,$ contrary to $G = \langle S \rangle.$ Hence $S = (St)t^{-1} \subseteq Mt^{-1}$ and $S \cap M = \emptyset$, since $t^{-1} \not \in M.$
On the other hand, if there is a maximal subgroup $M$ with $S \cap M = \emptyset,$ then for each $t \in S$, we have $S \subseteq Mt,$ since $G$ is the disjoint union $M \cup Mt.$ Then $St \subseteq Mt^{2} = M,$ so that $\langle St \rangle \neq G.$
As in @DaveBenson's answer, this only requires $G$ to be a finite $2$-group, not necessarily Abelian.
answered May 7 at 11:54