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Let $G$ be a group and $H < G$ a subgroup such that $H \subsetneq G$. Suppose $G$ is created by $H$ and another element $w \in G - H$ of order $2$, such that $wHw^{-1}=H$. Prove: $[G:H]=2$.


I tried to show that for every $g \notin H$ we have $wg \in H$, but had some trouble proving it. Any ideas of a proof (an elegant one if possible)?

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  • $\begingroup$ Try to prove that $wH$ and $H$ are the only two distinct left cosets of $H$ in $G.$ Observe that $G= \langle H,w \rangle.$ So the elements of $G$ are of the form $w^i h$ for $i=0,1,$ and $h \in H.$ $\endgroup$ – Dbchatto67 Apr 8 at 15:16
  • $\begingroup$ That's exactly what I have tried, that's why I want to show that if $g \notin H$ then $wg \in H$, but as I said I got stuck. $\endgroup$ – user401516 Apr 8 at 15:21
  • $\begingroup$ If $g \notin H$ then it is of the form $wh$ for some $h \in H.$ Therefore $wg = w (wh) =w^2 h = h \in H,$ as claimed. $\endgroup$ – Dbchatto67 Apr 8 at 15:24
  • $\begingroup$ Could you please explain why $g \notin H$ is necessarily of the form $wh, h \in H$? $\endgroup$ – user401516 Apr 8 at 15:27
  • $\begingroup$ Because every element of $G$ is of the form $w^ih$ for some $h \in H$ and where $i=0\ \text {or}\ 1.$ It could also be of the form $hw^i.$ But by the given condition $wHw^{-1} = H$ we can find out some element $h' \in H$ such that $hw^i=w^ih'.$ So we can conclude that every element of $G$ can be written by $w^ih$ for some $h \in H$ and $i=0$ or $1.$ Now if $i=0$ then the elements look like $h$ where $h \in H.$ Therefore in order to get an element in $G-H$ we should take the elements of the form $wh$ for some $h \in H.$ Is it clear now? $\endgroup$ – Dbchatto67 Apr 8 at 15:33
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Since $G$ is generated by $H$ and $G$, and $wHw^{-1} = H$, we can show that any element in the group is either an element of $H$, or it is of the form $hw$ for some $h\in H$. This means that $H$ has two cosets $H$ and $Hw$, and therefore $[G:H] = 2$.

I made a claim above. Let's show that. Take an element in $G$. Since $G$ is generated by $H$ and $w$, it can be written as some finite product on (at least one) one of the following four forms: $$ wh_1wh_2wh_3\cdots h_{n-1}wh_nw\\ wh_1wh_2wh_3\cdots h_{n-1}wh_n\\ h_1wh_2wh_3\cdots h_{n-1}wh_nw\\ h_1wh_2wh_3\cdots h_{n-1}wh_n $$ Consider what $wHw^{-1} = H$ means. It means that for any $h\in H$, there is a $h'\in H$ such that $$ whw^{-1} = h'\\ wh = h'w $$ Which is to say, any time in that product where we have a $w$ to the left of a $h_k$, we can move that $w$ to the other side of the $h_k$, as long as we change that $h_k$ to some corresponding $h'_k$.

This is something we can keep doing until all the $w$'s are to the right, and all the $h_i$'s are on the left, so our product becomes $$ h_1'h_2'\cdots h_n'ww\cdots w $$ That long product of $h_i'$'s results in a single element $h\in H$. And we know that $w$ has order $2$, so that long product of $w$'s is either going to end up being the identity, or just $w$. Which is to say, our arbitrary element of $G$ is of either the form $h$, or the form $hw$.

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  1. An alternative (but of course similar) argument would go as follows. Assume that you are given a group $G$ together with a proper subgroup $H<G$ and an element $a \in G$ of order $2$ such that $G=\langle H \cup \{a\} \rangle$ and $aHa^{-1}=H$. The glaring hypothesis that $a$ normalizes $H$ can be exploited to infer that since $H, \{a\} \subseteq \mathrm{N}_G(H)$ one has $\mathrm{N}_G(H) \geqslant G$ and thus $H \triangleleft G$, $H$ is a normal subgroup. Next, consider the quotient group $G/H$ and the canonical surjection $\sigma: G \to G/H$. If you apply $\sigma$ to the relation $G=\langle H \cup \{a\} \rangle$ you obtain $G/H=\langle \sigma(a) \rangle$ (bear in mind that whenever $f \in \mathrm{Hom}_{\mathrm{Gr}}(G, G')$ is a morphism, then for any $X \subseteq G$ it holds that $f(\langle X \rangle)=\langle f(X) \rangle$). By denoting $b=\sigma(a)$, since $a^2=1_G$ you automatically have $b^2=1_{G/H}$, so the quotient is generated by an element of order $1$ or $2$. As $H$ is proper, the quotient can not be trivial, so then $|G/H|=|G:H|=2$.

  2. We can bring forth yet another method of reasoning. We shall show that $\{1_G,\ a\}$ forms a complete and independent system of representatives for the left congruence modulo $H$. Since $H \cup \{a\}$ generates $G$ and $H$ is proper, we necessarily have $a \notin H$, or otherwise we would derive the contradiction $H=G$. Hence, $a\ {}_{H}\not \equiv 1_G$ and the system $\{1_G,\ a\}$ is indeed independent.

To prove the system is also complete we consider the subset $I=H \cup aH$ and show it is a left ideal of the semigroup $G$. As $I \neq \varnothing$ and a group only admits itself and the empty set as left ideals, it will follow right away that $I=G$.

Claiming that $I$ is a left ideal of $G$ amounts to establishing the inclusion $GI \subseteq I$. To this end define $M=\{t \in G\ |\ tI \subseteq I \}$, the left transporter of $I$ into itself. It is easy to verify that $M$ is a submonoid of $G$ and that $a \in M$ (this relies on the fact that $a^2=1_G$).

The crucial part is showing that $H \subseteq M$, but we do indeed have $HI=HH \cup HaH=H \cup (Ha)H=H \cup (aH)H=H \cup aH=I$. Agreeing to denote the submonoid generated by $X \subseteq G$ by $[X]$, we recall a remarkable lemma according to which $$\langle X \rangle=[X \cup X^{-1}]$$ As $a$ is of order $2$, the subset $H \cup \{a\}$ is symmetric (i.e. equal to its own inverse), by which we conclude - on the basis of the lemma above - that $$M \supseteq [H \cup \{a\}]=\langle H \cup \{a\} \rangle=G$$ hence $M=G$ and $I$ is indeed an ideal.

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