Is it the case that $Inn(G \times H) \cong Inn(G) \times Inn(H)$ where $G,H$ are some groups (not necessarily connected in any way)?

Question

The title basically. I was wondering if the inner automorphism group "distributed" over direct product. My intuition says yes but i am really not sure. My reasoning is that for any $\sigma_{(g,h)} \in Inn(G \times H)$ you can map it to $(\sigma_g,\sigma_h) \in Inn(G)\times Inn(H)$ and that would be an isomorphism.

Thanks in advance

Answer 1

Recall that $\operatorname{Inn}(G)\cong G/Z(G)$, where $Z(G)$ is the center of $G$. By fundamental properties of the center, we have that $Z(G\times H)=Z(G)\times Z(H)$ (see here). Also, by properties of the quotient group, if $N\trianglelefteq G$ and $K\trianglelefteq H$, then $(G\times H)/(N\times K)\cong G/N\times H/K$ (see here). Gathering everything together, we obtain $$\operatorname{Inn}(G\times H)\cong(G\times H)/(Z(G\times H))=(G\times H)/(Z(G)\times Z(H))\cong G/Z(G)\times H/Z(H)\cong\operatorname{Inn}(G)\times \operatorname{Inn}(H),$$ as desired. $\square$

Answer 2

While this follows from standard properties of the center and quotients, it's also easy to argue directly that your suggested map is an isomorphism. A benefit is that the implicit isomorphism becomes entirely explicit.

Via projections: One way to organize the argument is to first verify the following, each of which is simple:

1. $\sigma_{g_1} \circ \sigma_{g_2} = \sigma_{g_1 g_2}$
2. $\sigma_{(g,h)} = \sigma_g \times \sigma_h$
3. $\alpha \times \beta = \alpha' \times \beta'$ iff $\alpha = \alpha', \beta = \beta'$ (when the domains are non-empty)

Your suggested map is $\sigma_{(g, h)} \mapsto (\sigma_g, \sigma_h)$. Verify it's a well-defined bijection as follows: \begin{align*} \sigma_{(g, h)} = \sigma_{(g', h')} &\Leftrightarrow \sigma_g \times \sigma_h = \sigma_{g'} \times \sigma_{h'} \\ &\Leftrightarrow \sigma_g = \sigma_{g'}, \sigma_h = \sigma_{h'} \\ &\Leftrightarrow (\sigma_g, \sigma_h) = (\sigma_{g'}, \sigma_{h'}). \end{align*}

It's also a homomorphism, hence isomorphism: \begin{align*} \sigma_{(g_1, h_1)} \circ \sigma_{(g_2, h_2)} &= \sigma_{(g_1 g_2, h_1 h_2)} \\ &\mapsto (\sigma_{g_1 g_2}, \sigma_{h_1 h_2}) \\ &= (\sigma_{g_1} \circ \sigma_{g_2}, \sigma_{h_1} \circ \sigma_{h_2}) \\ &= (\sigma_{g_1}, \sigma_{h_1}) \circ (\sigma_{g_2}, \sigma_{h_2}). \end{align*}

Via centers: If you don't like products of maps, you can use centers instead:

1. $\sigma_g = \sigma_{g'}$ iff $g^{-1}g' \in Z(G)$
2. $(g, h) \in Z(G \times H)$ iff $g \in Z(G), h \in Z(H)$.

Well-defined bijection: \begin{align*} \sigma_{(g, h)} = \sigma_{(g', h')} &\Leftrightarrow (g, h)(g, h)^{-1} = (gg^{-1}, hh^{-1}) \in Z(G \times H) \\ &\Leftrightarrow gg^{-1} \in Z(G), hh^{-1} \in Z(H) \\ &\Leftrightarrow \sigma_g = \sigma_{g'}, \sigma_h = \sigma_{h'} \\ &\Leftrightarrow (\sigma_g, \sigma_h) = (\sigma_{g'}, \sigma_{h'}). \end{align*}

(The homomorphism verification is the same.)