Using a standard presentation of a semidirect product in combinatorial group theory, show $G\rtimes_{\operatorname{conj}}G\cong G\times G$.

Question

This is an alternative-proof question. I have a proof; I just want a different one using a particular technique. It's just for fun.

Using a standard presentation of a semidirect product in combinatorial group theory, show $$G\rtimes_{\operatorname{conj}}G\cong G\times G$$ for any group $G$.

Using standard combinatorial group theory,

$$\langle X\mid R\rangle \rtimes_\varphi \langle Y\mid S\rangle \cong \langle X\cup Y\mid R\cup S\cup \{\color{orange}{x^{y}}=\color{purple}{\varphi(y)(x)}\mid x\in X, y\in Y\}\rangle.$$

(For the question that inspired this one, see here.)

But $\operatorname{conj}(y)(x)=x^y$, so that

$$\color{orange}{x^y}=yxy^{-1}=\color{purple}{\operatorname{conj}(y)(x)}=x^y.$$

This is a tautology.

I was expecting $x^y=x$, which would render the desired result. (Here $x,y$ are from different (copies of a) generating sets (or set).)

At any rate, I would like Tietze transformations, say, to transform $G\rtimes_{\operatorname{conj}}G$ into $G\times G$.

I've twisted myself in knots trying to do what should be a simple exercise.

Answer 1

Let me just write down the Tietze transformation process:

Let $G = \langle X_G|R_G \rangle$. For clarity, take $X_H = X_G$ and $R_H = R_G$ so $H = G$. Also, for $y \in X_H$, write $y_G$ to mean the same element, but regarded as an element of $X_G$ instead. Note that, since $\varphi$ represents the conjugation action of $G$ on itself, we have, by definition, $\varphi(y)(x) = y_Gx(y_G)^{-1}$ for all $x \in X_G$, $y \in X_H$.

Now, $G \rtimes_\varphi G = G \rtimes_\varphi H$ can be presented by,

$$\langle X_G \cup X_H|R_G \cup R_H \cup \{yxy^{-1} = \varphi(y)(x)|x \in X_G, y \in X_H\}\rangle$$

Replacing $\varphi(y)(x)$ by its definition, we rewrite this as,

$$\langle X_G \cup X_H|R_G \cup R_H \cup \{yxy^{-1} = y_Gx(y_G)^{-1}|x \in X_G, y \in X_H\}\rangle$$

For each $y \in X_H$, add a new generator $y’ = (y_G)^{-1}y$,

$$\langle X_G \cup X_H \cup \{y’|y \in X_H\}|R_G \cup R_H \cup \{yxy^{-1} = y_Gx(y_G)^{-1}|x \in X_G, y \in X_H\} \cup \{y’ = (y_G)^{-1}y|y \in X_H\}\rangle$$

Since $y = y_Gy’$ follows from the relation $y’ = (y_G)^{-1}y$, and noting that $y_G \in X_G$ by definition, we may remove the generators in $X_H$,

$$\langle X_G \cup \{y’|y \in X_H\}|R_G \cup \{w(y_Gy’)|w \in R_H\} \cup \{y_Gy’x(y’)^{-1}(y_G)^{-1} = y_Gx(y_G)^{-1}|x \in X_G, y \in X_H\} \cup \{y’ = (y_G)^{-1}y_Gy’|y \in X_H\}\rangle$$

The relations $y’ = (y_G)^{-1}y_Gy’$ are trivial and can be removed. The relations $y_Gy’x(y’)^{-1}(y_G)^{-1} = y_Gx(y_G)^{-1}$ is easily seen to be equivalent to $y’x = xy’$, so,

$$\langle X_G \cup \{y’|y \in X_H\}|R_G \cup \{w(y_Gy’)|w \in R_H\} \cup \{y’x = xy’|x \in X_G, y \in X_H\}\rangle$$

Renaming $y’$ back to $y$,

$$\langle X_G \cup X_H|R_G \cup \{w(y_Gy)|w \in R_H\} \cup \{yx = xy|x \in X_G, y \in X_H\}\rangle$$

Under the relations $yx = xy$ (and recalling $y_G \in X_G$), inducting on words, we see that $w(y_Gy) = w(y_G)w(y)$. Then, under the relations in $R_G$ - which equals $R_H$ and so, recalling also that $X_G = X_H$, is equal to $\{w(y_G)|y \in X_H\}$, we see that $w(y_Gy) = 1$ is equivalent to $w(y) = 1$. Thus, we may replace $\{w(y_Gy)|w \in R_H\}$ by $\{w(y)|w \in R_H\} = R_H$.

$$\langle X_G \cup X_H|R_G \cup R_H \cup \{yx = xy|x \in X_G, y \in X_H\}\rangle$$

This is, of course, a presentation of $G \times H = G \times G$.