TITLE: Isaacs Algebra : A Graduate Course definition for diagonal subgroup
QUESTION [3 upvotes]: I actually have a number of related questions.
In Problem 7.2, Isaacs defines a subgroup $D$ of $G= M \mathop{\dot{\times}} N$
to be diagonal if
$$ D \cap M = 1 = D \cap N \text{ and } DM = G = DN. $$

Does Isaacs' definition for a diagonal subgroup change the usual definition much? ... where by usual definition of diagonal subgroup I mean the subgroup
$$ D = \{ (h, h) \mid h\in H \} \text{ of } G = H \times H. $$
(That's the definition I was aware of before, though that definition
can be extended to $G$ being an $n$-fold product of $H$, and $D$ consisting of $n$-tuples of the same element of $H$.)

In Problem 7.2, one proves a diagonal subgroup $D$ of $G$ exists iff $M\cong N$.
Does $D$ diagonal in $M \mathop{\dot{\times}} N$ effectively mean
$$ D = \{ m\cdot m\varphi \mid m\in M\}, \text{ where } N = M\varphi ?$$

Are there other sources discussing Isaacs' definition?

If Isaacs' definition is essentially equivalent to the usual definition,
is there a way to nicely write the proof of Problem 7.7:
With $G, M, N, D$ defined as above, prove
$$M \text{ (and therefore $N$) simple } \implies D \text{ is a maximal subgroup of }G.$$
so that it reduces to the solution given in The Diagonal Subgroup of $A \times A$ is Maximal iff $A$ is Simple ?
Seemingly related is Showing that the diagonal of $G \times G$ is maximal, where $G$ is simple .

REPLY [3 votes]: Isaacs's definition of diagonal subgroup is invariant under automorphisms of one of the factors, whereas the 'standard' definition is not. Basically, if $A$ and $B$ are isomorphic groups then any subgroup of $A\times B$, isomorphic to $A$, and sitting diagonally between them (so the intersections with each term are trivial) could be called a diagonal. You are thinking of the special case $B=A$, rather than just $B\cong A$. If you have two groups that are just isomorphic, different identifications of the two groups will yield different diagonal subgroups with your definition. Isaacs's allows for the second factor to be twisted by an automorphism when defining the subgroup.

Yes, these are often called twisted diagonal subgroups. That the two definitions are equivalent is a fairly easy exercise (apparently it's 7.2).

Not many. This is all you really need to know. Every twisted diagonal subgroup is the diagonal subgroup in your sense, with an identification of the two groups being through an automorphism of one factor. Somebody else on this website also got confused about this point, and has been left unhelped for the last six years.

Well, given that twisted diagonal subgroups are just the closure under automorphisms of one of the factors of 'standard' diagonal subgroups, any proof should follow through.
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