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I am reading a book about Representation Theory of Finite Groups and in the book it gives:

$End(W) \cong M_n(\mathbb{C})$
$GL(V) \cong GL_n(\mathbb{C})$
$Hom(V,W) \cong M_{mn}(\mathbb{C})$

$V$ and $W$ are Vector Spaces and $Hom(V,W) = \{A:V\rightarrow W | A$ is a linear map$\}$, $End(V) = Hom(V,V)$ and $GL(V) = \{A \in End(V)|A$ is invertible$\}$

I understand what these sets are, but in the above relations I was unsure whether they are being considered as groups (With group operation composition of maps/multiplication of matrices) or as vector spaces (with addition of maps/matrices and scalar multiplication by scalars in $\mathbb{C}$).

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  • $\begingroup$ Which book are you referring to? $\endgroup$
    – Shaun
    Oct 26, 2021 at 20:17
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    $\begingroup$ "Benjamin Steinberg - Representation Theory of Finite Groups: An Introductory Approach" $\endgroup$
    – 123123
    Oct 26, 2021 at 20:19

1 Answer 1

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$\operatorname{End}(W) \cong M_n(\mathbb{C})$ as $\mathbb{C}$-algebras (i.e. as rings as well as $\mathbb{C}$-vector spaces), $\operatorname{GL}(V) \cong \operatorname{GL}_n(\mathbb{C})$ as groups, and $\operatorname{Hom}(V,W) \cong M_{mn}(\mathbb{C})$ as $\mathbb{C}$-vector spaces.

EDIT

Note that $\operatorname{End}(W)$ is not a group wrt composition (as it contains non-invertible elements) and $\operatorname{GL(V)}$ is not a vector space (as it doesn't contain an additive zero element, for example). So there is no way to read all three $\cong$ signs uniformly as "isomorphic as vector spaces" nor uniformly as "isomorphic as multiplicative groups".

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  • $\begingroup$ I would have thought they would all be either groups or vector spaces since the three relations are given together. $\endgroup$
    – 123123
    Oct 26, 2021 at 20:18
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    $\begingroup$ @123123 No, that's not possible. $\operatorname{End}(W)$ is not a group, and $\operatorname{GL}(V)$ is not a vector space, so no way to read all $\cong$ signs as "isomorphic as gropus" nor as "isomorphic as vector spaces". $\endgroup$
    – azimut
    Oct 26, 2021 at 20:22
  • $\begingroup$ Right, because the linear map would need to be invertible to be a group! And GL(V) is not a vector space because the sum of two invertible matrices is not necessarily invertible. Thank you. $\endgroup$
    – 123123
    Oct 26, 2021 at 20:27
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    $\begingroup$ $End(V)$ is not a group under multiplication. As a vector space, or as a ring, it is always a group under addition. $\endgroup$ Oct 26, 2021 at 20:51
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    $\begingroup$ @ThomasAndrews well...you are right of course. "isomorphic as vector spaces" implies "isomorphic as (additive) groups", so what I wrote is not strictly true. But I guess that "as a group" referred to the multiplicative structure. I guess I will remove the addition from my answer again. $\endgroup$
    – azimut
    Oct 26, 2021 at 21:00

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