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Let $G$ be a finite group and $M$ is a maximal subgroup of $G$. Prove $\forall g \in G$, $M^{g}$ is a maximal subgroup of $G$.

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If $N$ is a subgroup of $G$ properly containing $M^g$, then $N^{g^{-1}}$ … –  egreg Jun 9 '13 at 20:17

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up vote 3 down vote accepted

The map $x\mapsto x^g$ is an automorphism of $G$. Its inverse is $x\mapsto x^{g^{-1}}$. What is $x^g$ depends on conventions, but your notation seems to imply that $$ x^g=g^{-1}xg $$ so that, for $g,h\in G$, $$ (x^g)^h=h^{-1}(x^g)h=h^{-1}g^{-1}xgh=x^{gh}. $$ More generally, if $\varphi$ is an automorphism of $G$ (that is, a bijective homomorphism $\varphi\colon G\to G$), when $M$ is a subgroup of $G$, also $M^\varphi=\{\,x^g:x\in M\,\}$ is a subgroup of $G$. Also $$ M\subseteq N \to M^\varphi\subseteq N^\varphi $$ where $M$ and $N$ are arbitrary subsets of $G$, in particular subgroups. If $\psi$ denotes the inverse automorphism of $\varphi$ (just to avoid stacked superscripts), we can also consider the obvious fact that $(M^\varphi)^\psi=M$, so that the map $M\mapsto M^\varphi$ is an inclusion preserving bijection in the set of subgroups of $G$.

Let $M$ be a maximal subgroup of $G$. Suppose $N$ is a subgroup of $G$ properly containing $M^\varphi$. Then $N^\psi$ properly contains $M=(M^\phi)^\psi$ and, by maximality, $N^\psi=G$. Therefore $N=(N^\psi)^\varphi=G^\varphi=G$.

Note that the finiteness of $G$ is not needed. If maximality is understood in terms of “maximal cardinality”, the assertion is pretty obvious, as $|M|=|M^\varphi|$.

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you can interpret the conjugation operator in the same way as a change of basis in linear algebra (some people go as far as to call change of basis conjugation). thus one's intuition when talking about conjugation should be (roughly) that properties of a subgroup should remain more-or-less unchanged after it –  citedcorpse Jun 10 '13 at 11:24

The map $f_g : G \to G$ given by $f_g(h) = ghg^{-1}$ is a special kind of automorphism called an "innermorphism" (or inner automorphism). Can you see how this answers your question?

EDIT — a somewhat subtle point brought to my attention by Julian: we must distinguish between "subgroup not contained in any proper subgroup" (i.e. a maximal subgroup) and "subgroup of maximal cardinality". These two notions do not necessarily coincide: for instance, in $S_3$ we have two kinds of proper subgroups. One of these is $A_3$, and the other is generated by any $2$-cycle. Both are clearly maximal, yet only $A_3$ has maximal cardinality.

Thus the correct way to view this is as the comment on the original post does, and not the one based on cardinality: any would-be subgroup containing $M^g$ can be conjugated to contain $M$, and hence must either be $M$ or $G$. Thus $M^g$ is also maximal.

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yes, In fact |$M$|=|$M^{g}$| –  M.Mazoo Jun 9 '13 at 20:37

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