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My lecturer wrote that a matrix representation of a group $G$ is a group homomorphism $\rho$ from $G$ to the set of invertible square matrices over a field. He proceeded to write that this is equivalent to the following two conditions:

  1. $\rho(\varepsilon)=I$ (where $\varepsilon$ is the identity in $G$ and $I$ the identity matrix).
  2. $\rho(gh)=\rho(g)\rho(h)$ for all $g,h\in G$.

But am I wrong to say that condition 1 follows directly from condition 2? Because condition 2 tells us that $$ \rho(g)=\rho(\varepsilon g)=\rho(\varepsilon)\rho(g) $$ for any $g\in G$, which implies that $\rho(\varepsilon)$ is the identity matrix. So why include condition 1?

Update: In the comments the doubt above was solved. Now I would like to actually show that $\rho=I$ is equivalent to saying that the homomorphism $\rho$ maps all $g\in G$ to the set of invertible square matrices. My intend:

  1. If $\rho(\varepsilon)=I$ then $I=\rho(\epsilon)=\rho\left(g g^{-1}\right)=\rho(g) \rho\left(g^{-1}\right)$, which shows that $\rho(g)$ has an inverse given by $\rho(g)^{-1}=\rho(g^{-1}$).
  2. If we know that $\rho(\varepsilon)$ has an inverse then $$ \rho(\varepsilon)\rho(\varepsilon)=\rho(\varepsilon\varepsilon)=\rho(\varepsilon)=I\rho(\varepsilon)\iff\rho(\varepsilon)\rho(\varepsilon)\rho(\varepsilon)^{-1}=I\rho(\varepsilon)\rho(\varepsilon)^{-1} \iff\rho(\varepsilon)=I. $$ Is there anything wrong with this proof?
Manó
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    You are indeed wrong - for example, $\rho$ could just map everything to the zero matrix. – Izaak van Dongen Mar 20 '23 at 17:12
  • Ah, right. I did not think of that. Thank you. – Manó Mar 20 '23 at 17:19
  • @IzaakvanDongen except zero matrix is not invertible. OP is correct. This works exactly the same as with normal group homomorphisms. In fact representation is just a group homomorphism from $G$ to $GL_n(k)$. – freakish Mar 20 '23 at 17:21
  • @freakish, My understanding is that OP thought that condition 2 implies condition 1. My example shows it doesn't. The point is that (1) and (2) characterise "is a homomorphism to the group of invertible matrices", but (2) by itself doesn't. Conditions (1) and (2) are implicitly in the context "$\rho$ is a map from $G$ to the set of matrices", usually. – Izaak van Dongen Mar 20 '23 at 17:24
  • @IzaakvanDongen yeah, but OP explicitly stated "to the set of invertible square matrices". The devil always is in the details. – freakish Mar 20 '23 at 17:26
  • @freakish, from what I understand (now) condition 1 is there to make sure that $\rho(g)$ is invertible. As Izaak writes, to me it seemed that condition 1 followed from condition 2, which is not always the case by his example. – Manó Mar 20 '23 at 17:26
  • @Manó you explicitly wrote "to the set of invertible square matrices". Which is a correct and important assumption. In this context, 1 follows from 2. – freakish Mar 20 '23 at 17:27
  • Yes, but then he says that this is equivalent to the two conditions below. But condition two by itself is not equivalent to saying that $\rho(g)$ is invertible. – Manó Mar 20 '23 at 17:28
  • @Manó well, those statements are meaningless to begin with, if you don't state what domain and range are. So we can safely assume that the domain and range are common for both statements. Meaning the author simply wrote a definition of homomorphism in a confusing way. – freakish Mar 20 '23 at 17:31
  • @freakish, when I learned representation theory I saw a remark a lot like this, and I think the intent is almost certainly to say "a map $\rho: G \to \mathrm{Mat}_n(K)$ is a representation iff (1) and (2) hold" - which is certainly the most interesting possible interpretation of this remark. Perhaps it was worded a little unclearly, or something was lost in transcription. Checking that the matrices you define are actually invertible is an annoying thing you have to do in representation theory that a lot of people who already know group theory forget to do, which is why this remark is here. – Izaak van Dongen Mar 20 '23 at 17:32
  • @IzaakvanDongen well, maybe we read different books, because I've never seen a representation defined as a function to $Mat_n(K)$. But sure, if someone does this then you are correct. – freakish Mar 20 '23 at 17:34
  • This seems relevant: https://math.stackexchange.com/questions/1161115/a-representation-theory-problem-in-putnam-competition/ . Explicitly saying that $\rho(e)=I$ allows you to confirm that determinant $\rho(g)$ is a unit (i.e. non-zero since we have a field) for all $g$. Incidentally $\det\big(\rho(g)\big)$ gives another representation [if you identify $1\times 1$ matrices with scalars]. – user8675309 Mar 20 '23 at 23:46

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