Problem 5.2.13 (Presentation of a direct product)

Question

Let $A = \langle x_1 angle  	imes \cdots 	imes \langle x_r angle $ be a finite abelian group with $|x_i| = n_i$ for $1 \leq i \leq r$. Find a presentation for $A$. Prove that if $G$ is any group containing commuting elements $g_1,g_2,\ldots,g_r$ such that $g_i^{n_i} = 1$ for $1 \leq i \leq r$, then there is a unique homomorphism from $A$ to $G$ which sends $x_i$ to $g_i$ for all $i$.

Richard Ganaye · Accepted Answer

\begin{proof} Consider the group $\Gamma$ defined by
$$\Gamma = \langle X_1,\ldots,X_r \mid X_1^{n_1} = \cdots = X_r^{n_r} = 1, \ X_i X_j = X_j X_i \qquad (1\leq i \leq n, \ 1 \leq j \leq n) angle.$$
We identify $x_1$ with $(x_1, 1, \ldots,1) \in A$, and so on. Since $x_1^{n_1} = x_2^{n_2} = \cdots = x_r^{n_r} = 1$ and $x_i x_j = x_j = x_i$ for all $i,j$, where $A = \langle x_1,\ldots,x_rangle$,  there is a surjective homomorphism
$$\varphi : \Gamma 	o A,$$
such that $\varphi(X_i) =x_i$ for all $i$.

Since the $X_i$ are commuting element, every $X$ element of $\Gamma$ is of the form
$$X = X_1^{k_1}\cdots X_r^{k_r},\qquad 	ext{where} \qquad 0\leq k_i < n_i\ (1 \leq i \leq n).$$

Therefore $|\Gamma| \leq n_1\cdots n_r$. Since $\varphi$ is surjective, $$n_1\cdots n_r = |A| \leq |\Gamma| \leq n_1\cdots n_r,$$
thus $|A| = |\Gamma|$. Therefore $\varphi$ is an isomorphism, and
\begin{align*}
A &= \langle x_1 angle  	imes \cdots 	imes \langle x_r angle \
&\simeq \langle X_1,\ldots,X_r \mid X_1^{n_1} = \cdots = X_r^{n_r} = 1, \ X_i X_j = X_j X_i \qquad (1\leq i \leq n, \ 1 \leq j \leq n) angle.
\end{align*}
Changing notations, we can write
\begin{align*}
A &= \langle x_1 angle  	imes \cdots 	imes \langle x_r angle\
&\simeq \langle x_1,\ldots,x_r \mid x_1^{n_1} = \cdots = x_r^{n_r} = 1, \ x_i x_j = x_j x_i \qquad (1\leq i \leq n, \ 1 \leq j \leq n) angle.
\end{align*}

\bigskip
If $G$ is any group containing commuting elements $g_1,g_2,\ldots,g_r$ such that $g_i^{n_i} = 1$ for $1 \leq i \leq r$, then the elements $g_i$ satisfy the relations of $\Gamma$, so there is an homomorphism  from $A\simeq \Gamma$ to $G$ which sends $x_i$ to $g_i$ for all $i$.
\end{proof}

Problem 5.2.13 (Presentation of a direct product)

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Problem 5.2.13 (Presentation of a direct product)

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