If you find any mistakes, please make a comment! Thank you.

Basic properties of n-ary direct products of groups


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 5.1 Exercise 5.1.16

Solution:

(1) First we show that $\iota_k$ is an injective homomorphism. If $g,h \in G_k$, then $(\iota_k(gh))_i = gh$ if $i=k$ and 1 otherwise. In either case, we have $$(\iota_k(gh))_i = (\iota_k(g))_i (\iota_k(h))_i.$$ Thus $\iota_k$ is a homomorphism. Now suppose $g \in \mathsf{ker}\ \iota_k$. Then $\iota_k(g) = \prod 1_i$, so that in particular $g = 1$. Thus $\mathsf{ker}\ \iota_k = 1$, hence $\iota_k$ is injective.

Now suppose $(\prod g_i) \in G$ and $(\prod a_i) \in \mathsf{im}\ \iota_k$. Now $$(\prod g_i)(\prod a_i)(\prod g_i)^{-1} = \prod (g_ia_ig_i^{-1}),$$ and if $i \neq k$, we have $g_ia_ig_i^{-1} = 1$. Then $$(\prod g_i)(\prod a_i)(\prod g_i)^{-1} = \iota_k(g_ka_kg_k^{-1}),$$ hence $\mathsf{im}\ \iota_k$ is normal in $G$.

Now define a mapping $$\theta : G \rightarrow \prod_{I \setminus \{k\}} G_i by (\theta(\prod g_i))_j = g_j.$$ Now for all $j \in I \setminus \{k\}$, $$(\theta((\prod g_i)(\prod h_i)))_j = (\theta(\prod g_ih_i))_j = g_jh_j = (\theta(\prod g_i))_j (\theta(\prod h_i))_j,$$ so that $\theta$ is a group homomorphism. If $\prod h_i \in \prod_{I \setminus \{k\}} G_i$, then $\prod g_i \in G$ defined by $g_i = h_i$ if $i \neq k$ and $g_k = 1$ maps to $\prod h_i$ under $\theta$; hence $\theta$ is surjective. Finally, we see that $\mathsf{ker}\ \theta = \mathsf{im}\ \iota_k$ as follows.

($\subseteq$) If $\theta(\prod g_i) = 1$, then $g_i = 1$ for all $i \neq k$. Thus $\prod g_i = \iota_k(g_k)$.

($\supseteq$) If $g \in G_k$, then $(\theta(\iota_k(g)))_j = 1$ for all $j \neq k$; hence $\iota_k(g) \in \mathsf{ker}\ \theta$. Finally, by the First Isomorphism Theorem, we have $$G/\mathsf{im}\ \iota_k \cong \prod_{I \setminus \{k\}} G_i.$$

(2) First we show that $\pi_k$ is a surjective homomorphism. We have $$\pi_k((\prod g_i)(\prod h_i)) = \pi_k(\prod g_ih_i) = g_kh_k = \pi_k(\prod g_i) \pi_k(\prod h_i).$$ Thus $\pi_k$ is a homomorphism. Now let $h \in G_i$, and define $\prod g_i \in G$ by $g_i = h$ if $i = k$ and 1 otherwise. Clearly then $\pi_k(\prod g_i) = h$; hence $\pi_k$ is surjective.

Now define $\tau : \mathsf{ker}\ \pi_k \rightarrow \prod_{I \setminus \{k\}} G_i$ as follows: $(\tau(\prod g_i))_j = g_j$. This $\tau$ is clearly a group homomorphism. Now if $\prod h_i \in \prod_{I \setminus \{k\}} G_i$, then $\prod g_i \in G$ defined by $g_i = 1$ if $i=k$ and $h_i$ otherwise is in $\mathsf{ker}\ \pi_k$, and moreover $\tau(\prod g_i) = \prod h_i$. Thus $\tau$ is surjective. Finally, if $\prod g_i \in \mathsf{ker}\ \tau$, then $g_i = 1$ for all $i \neq k$, and (by definition) $g_k = 1$. Thus $\prod g_i = 1$, so that $\tau$ is injective. Thus $\mathsf{ker}\ \pi_k \cong \prod_{I \setminus\{k\}} G_i$.

(3) Let $g \in G_i$ and $h \in G_j$ where $i \neq j$. Then $(\iota_i(g) \iota_j(h))_k$ is $g$ if $k=i$, $h$ if $k=j$, and is 1 otherwise. Similarly for $(\iota_j(h) \iota_i(g))_k$; thus these elements are equal.


Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

Leave a Reply

Close Menu