**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.