# Group generated by a proper subgroup and an element of order $2$ has index $2$

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)?

• 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 $h \in H.$ – Dbchatto67 Apr 8 at 15:16
• 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. – user401516 Apr 8 at 15:21
• 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. – Dbchatto67 Apr 8 at 15:24
• Could you please explain why $g \notin H$ is necessarily of the form $wh, h \in H$? – user401516 Apr 8 at 15:27
• 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? – Dbchatto67 Apr 8 at 15:33

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 $$hw$$.

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 $$H and an element $$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.