Show that $S=\{ A\in GL_n(K)\mid AJA^t=J\}$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.

Question

I have a question about the following problem: show that $$S=\{ A \in GL_n(K) \mid AJA^t = J \}$$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.

I have shown that the identity element $e$ is part of $S$, but I don't know to prove the second criteria for a subgroup. Thanks in advance!

It might be worth noting that this is a special case of the general principle that the stabilizer of any element under a group action is a subgroup of the acting group. In this case, $GL_n(K)$ acts on $K^{n \times n}$ via $A \cdot J = A J A^t$. — Daniel Schepler, Feb 09 '21 at 00:03
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score 2 · Answer 1 · answered Feb 08 '21 at 21:51

2

Assume that $A,B \in S$. This means $A,B \in GL_n(K)$ and $AJA^t = BJB^t = J$. We wish to show $AB \in S$. We have $AB \in GL_n(K)$ and $$(AB)J(AB)^t = (AB)J(B^tA^t) = A(BJB^t)A^t = AJA^t = J$$ so indeed $AB \in S$.

answered Feb 08 '21 at 21:51

mechanodroid

47,570

I feel really dumb... thanks again! – PDP Feb 08 '21 at 21:53
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You still need to show that the inverse of each element is in the subgroup. – Shaun Feb 08 '21 at 22:13
Each element of the proposed subgroup, that is . . . – Shaun Feb 08 '21 at 22:34

Shaun · Accepted Answer · 2021-02-08T23:42:03.973

1

I will use the one-step subgroup test.

Clearly $\varnothing\neq S\subseteq GL_n(K)$.

Let $A,B\in S.$ Then $AJA^t=J=BJB^t$. Observe that $$\begin{align} B^{-1}J(B^t)^{-1}&=B^{-1}BJB^t(B^t)^{-1}\\ &=J. \end{align}$$

We aim to show that $AB^{-1}\in S$. Indeed,

$$\begin{align} (AB^{-1})J(AB^{-1})^t&=(AB^{-1})J((B^{-1})^tA^t)\\ &=A(B^{-1}J(B^t)^{-1})A^t\tag{1}\\ &=AJA^t\\ &=J. \end{align}$$

Hence $S\le GL_n(K)$.

$(1)$: See here.

edited Feb 08 '21 at 23:42

answered Feb 08 '21 at 22:32

Shaun

47,747

1

Thanks! I understood the proof! – PDP Feb 09 '21 at 10:58

Show that $S=\{ A\in GL_n(K)\mid AJA^t=J\}$ is a subgroup of $GL_n(K)$, where $J\in K^{n\times n}$.

2 Answers2