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Why is $\langle a,b|bab^{-1} a^{-1},a a \rangle$ isomorphic to $\Bbb Z_2 \oplus \Bbb Z$?

Shaun
  • 44,997
Christian
  • 41

2 Answers2

4

Here's some observations.

The first relation tells u $a,b$ commute . The second relarion tells u $a^2=id$ So u have an abelian group $G$ generated by $<a,b>$ such that $a^2=id$

Consider the surjective $\mathbb Z-module $ homomorphism $$\mathbb Z \oplus \mathbb Z \rightarrow G$$ given by $$(1,0) \mapsto b$$ $$(0,1) \mapsto a$$

Compute the kernel of this and you will get $Ker = \mathbb Z.(0,2)$ Thus $$G\cong \frac {\mathbb Z \oplus \mathbb Z}{\mathbb Z. (0,2)} \cong \mathbb Z\oplus \mathbb Z_2$$

Problems like this can also be solved using algebraic topology. Consider $S^1$ with CW structure as $*\subset S^1$ and $\mathbb R\mathbb P^2$ with CW structure $*\subset S^1\cong \mathbb R \mathbb P^1 \subset \mathbb R \mathbb P^2$.

Now look at $X = S^1 \times \mathbb R \mathbb P^2$ $$\pi_1(X)= \mathbb Z \oplus \mathbb Z_2$$

But you can also compute the fundamental group explicitly from the product CW structure on $X$ . The product CW structure has $one \ 0-cell$, $ two \ 1-cell's$ and $two \ 2-cells$. The $1th \ skeleton$ is a wedge of two circles labelled $a,b$ The 1st $2-cell$ is pasted along $aa$ corresponding to the $(2-cell \ \ of \mathbb R \mathbb P^2 \ )\times (0-cell \ of \ S^1)$ and the other $2-cell$ is pasted along $bab^{-1}a^{-1}$. Then Van Kampens theorem gives u $$\pi_1(X)=<a,b|bab^{-1}a^{-1} ,aa>$$

This gives u a purely topological proof

user6
  • 4,092
0

Because $b$ is free besides the commutator $[b,a]=bab^{-1}a^{-1}$ being the identity, $aa=a^2=e$ means $a$ has order two, and those are the only relators. Hence

\begin{align} \Bbb Z_2\oplus\Bbb Z&\cong \langle a\mid a^2\rangle\oplus\langle b\mid\rangle \\ &\cong\langle a, b\mid a^2, [b, a]\rangle \\ &\cong \langle a, b\mid bab^{-1}a^{-1}, aa\rangle \end{align}

as desired.

Shaun
  • 44,997