Determining Whether Two Groups are Isomorphic by their Group Pres.

# Determining Whether Two Groups are Isomorphic by their Group Presentations

Definition: Let $G$ be a group and let $S = \{ X_1, X_2, ..., X_n \}$. We denote $W(X_1, X_2, ..., X_n)$ to be a word over $S \cup S^{-}$. Then we define the following set by $\{ W(X_1, X_2, ..., X_n) \}_G = \{ W(g_1, g_2, ..., g_n) : g_1, g_2, ..., g_n \in G \}$. |

We now prove a very important

Theorem 1: Let $G = \langle a_1, a_2, ... : P_1, P_2, ... \rangle$ and let $H = \langle b_1, b_2, ... : Q_1, Q_2, ... \rangle$. If $G$ and $H$ are isomorphic then for every $W(X_1, X_2, ..., X_n)$ we have that $\langle a_1, a_2, ... : P_1, P_2, ..., \{ W(X_1, X_2, ..., X_n) \}_G \rangle$ and $\langle b_1, b_2, ... : Q_1, Q_2, ..., \{ W(X_1, X_2, ..., X_n) \}_H \rangle$ are isomorphic. |

We now give an example of applying Theorem 1

## Example 1

**Prove that $G = \langle a, b : \emptyset \rangle$ is not isomorphic to $H = \langle x, y, z : \emptyset \rangle$.**

Suppose that $G$ and $H$ are instead isomorphic and let $S = \{ X, Y \}$. Let $W(X, Y) = XYX^{-1}Y{^-1}$. Then by Theorem 1 we have that:

(1)\begin{align} \quad \langle a, b : \{ W(X, Y) \}_G \rangle \cong \langle x, y, z : \{ W(X, Y) \}_H \rangle \end{align}

Observe that:

(2)\begin{align} \quad \langle a, b : \{ W(X, Y) \}_G \rangle &= \langle a, b : \{ XYX^{-1}Y^{-1} \}_G \rangle \\ &= \langle a, b : aba^{-1}b^{-1} = 1 \} \\ &= \langle a, b : ab = ba \} \\ &= \mathbb{Z} \times \mathbb{Z} \end{align}

(3)
\begin{align} \quad \langle x, y, z : \{ W(X, Y) \}_H \rangle &= \langle x, y, z : \{ XYX^{-1}Y^{-1} \}_G \rangle \\ &= \langle x, y, z : xyx^{-1}y^{-1} = 1, xzx^{-1}z^{-1} = 1, yzy^{-1}z^{-1} = 1 \rangle \\ &= \langle x, y, z : xy = yx, xz = zx, yz = zy \rangle \\ &= \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} \end{align}

Clearly $\mathbb{Z} \times \mathbb{Z} \not \cong \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$ and so the assumption that $G$ and $H$ are isomorphic was false.