# In $G_1 \times G_2$, $C(H_1 \times H_2) = C(H_1) \times C(H_2)$

Let $$H_1 \leq G_1$$ and $$H_2 \leq G_2$$. Show that in $$G_1 \times G_2$$, $$C(H_1 \times H_2) = C(H_1) \times C(H_2)$$

What I have done:

Here, $$C(X)$$ denotes the centralizer of $$X$$.

We have that $$C(H_1 \times H_2)=\{(g_1,g_2)\in G_1 \times G_2 : (g_1,g_2)(h_1,h_2)=(h_1,h_2)(g_1,g_2), (h_1,h_2)\in (H_1\times H_2)\}$$ and $$C(H_1) \times C(H_2)=\{(g_1,g_2)\in G_1 \times G_2:g_1h_1=h_1g_1, g_2h_2=h_2g_2 , h_i\in H_i \}$$, Which are the same set.Is this enough for my proof or do I need something else? Is this the way?

• Your notation is missing a very important thing which is where you're taking the centralisers. But yes, the proof that $C_{G_1}(H_1) \times C_{G_2}(H_2) = C_{G_1 \times G_2}(H_1 \times H_2)$ is essentially correct. May 25, 2021 at 17:41

The picture will be clean if you look: The product operation in $$G_1\times G_2$$ is pointwise.
In particular, $$(x,y)$$ centralizes $$H_1\times H_2$$ if and only if $$x$$ centralizes $$H_1$$ and $$y$$ centralizes $$H_2$$.