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Let $G =$ PGL$_{n}(\textbf{C})$ and $T$ be the image in $G$ of the subgroup of the invertible diagonal matrices of $\operatorname{GL}_{n}(\textbf{C})$. Let $A$ and $B$ be two elementary abelian $2$-subgroups in $T$ of the same rank.

The Kronecker product $\otimes I_{2}$ embeds $A$ and $B$ in $H$ = $\operatorname{PGL}_{2n}(\textbf{C})$. If $A$ and $B$ are not conjugate in $G$, will $A\otimes I_{2}$ and $B\otimes I_{2}$ not be conjugate in $H$? Intuitively, they are not conjugate in $H$ but I'm not sure of tools to tackle it. Maybe subgroup conjugacy is an equivalent relation and $\otimes I_{2}$ reserves it?

Edit: Some thinking made based on the input in the comment. If $A$ and $B$ are not conjugate in $G$, and we assume $A\otimes I_{2}$ and $B\otimes I_{2}$ are conjugate, attempt to get a contradiction. Since $A\otimes I_{2}$ and $B\otimes I_{2}$ are both block diagonal matrices with blocks $I_{2}$ and $-I_{2}$, if there always exists a "block permutation matrix" with each block $I_{2}$, then we're done. But I'm not sure about the existence of such a "block permutation matrix"... Any hints would be appreciated.

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    $\begingroup$ Here is your tool: Two diagonal matrices are conjugate to each other if and only if the elements on the diagonal are permutations of each other (because these are the eigenvalues of these matrices, hence invariant under conjugation). $\endgroup$ Commented Nov 8, 2022 at 10:15

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If $A$ and $B$ are elementary abelian $2$-subgroups of $\mathrm{PGL}_n(\mathbf C)$ of rank $r$ then they lift uniquely to elementary abelian $2$-subgroups of $\mathrm{GL}_n(\mathbf C)$ of rank $r+1$ (take all lifts $a$ of elements of $A$ such that $a^2 = 1$), so let us assume $A, B \le \mathrm {GL}_n(\mathbf C)$ to begin with.

Consider a representation $\rho:C_2^r \to \mathrm{GL}_n(\mathbf C)$ with character $\chi$. Decompose $\chi$ into irreducible characters: $\chi = w_1 \chi_1 + \cdots + w_m \chi_m$. Here $\chi_1, \dots, \chi_m : C_2^r \to \{\pm1\}$ and $w_1, \dots, w_m$ are positive integers. Two representations $\rho$ and $\rho'$ are equivalent (conjugate) if and only if they have the same character, so equivalence classes of representations $\rho:C_2^r \to \mathrm{GL}_n(\mathbf C)$ are in bijection with maps $w: D \to \mathbf N = \{0,1,2,\dots\}$ such that $\sum_{\chi \in D} w(\chi) = n$, where $D \cong C_2^r$ is the dual group of $C_2^r$ (the group of characters $\chi : C_2^r \to \{\pm1\}$ with pointwise product as the group operation). Here $w(\chi_i)$ records the multiplicity of an irreducible character $\chi_i$ in $\chi$. The representation is faithful if and only if the support $\mathrm{supp}(w) = \{\chi : w(\chi) > 0\}$ of $w$ generates $D$.

To get conjugacy classes of embedded copies of $C_2^r$ we must further quotient $\{w:D \to \mathbf N \mid \langle \mathrm{supp}~w\rangle = D\}$ by the action of $\mathrm{Aut}(C_2^r) \cong \mathrm{GL}_r(2)$.

In any case the Kronecker product just sends $w$ to $2w$, so the conjugacy relation is preserved.

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  • $\begingroup$ I really appreciate your input. May I ask about the definition of the dual of a group and the support of a map? Also, I'm not quite sure how $w$ maps elements... And how do we sum up the $w$ so the sum is $n$? Sorry for the trouble. Thank you! $\endgroup$
    – user488802
    Commented Nov 10, 2022 at 10:50
  • $\begingroup$ I added some words. I hope it helps. $\endgroup$ Commented Nov 10, 2022 at 11:48
  • $\begingroup$ Thank you for the added words which helped me a lot. Could you please point me to a proof of "the representation is faithful iff the support of $w$ generates $D$"? $\endgroup$
    – user488802
    Commented Nov 11, 2022 at 9:12
  • $\begingroup$ Also, I don't quite get this: to get the conjugacy classes in $\operatorname{PGL}_{n}(\textbf{C})$, we quotient $\{w:D \to \mathbf N \mid \langle \mathrm{supp}~w\rangle = D\}$ by the action of $\mathrm{Aut}(C_2^r) \cong \mathrm{GL}_r(2)$. Could you please elaborate a bit more? Thank you very much! $\endgroup$
    – user488802
    Commented Nov 11, 2022 at 9:16
  • $\begingroup$ For the first: if $\chi = w_1 \chi_1 + \cdots + w_m \chi_m$ then $\rho$ is equivalent up to conjugation to a diagonal representation with $\chi_1, \dots, \chi_m$ entries (and the appropriate multiplicities). Clearly this is faithful iff $\ker \chi_1 \cap \cdots \cap \ker \chi_m = 0$, which is equivalent to $\mathrm{supp}(w)$ generating $D$. $\endgroup$ Commented Nov 11, 2022 at 10:04

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