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?

  • 2
    $\begingroup$ 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. $\endgroup$ May 25, 2021 at 17:41

1 Answer 1


The picture will be clean if you look: The product operation in $G_1\times G_2$ is pointwise.

Thus, for defining centralizer of some element or some subgroup, you need to consider it pointwise.

In particular, $(x,y)$ centralizes $H_1\times H_2$ if and only if $x$ centralizes $H_1$ and $y$ centralizes $H_2$.

This is nothing but your statement.


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