Why is $\Omega (M)$ a superfluous submodule? I'm struggling with a result that seems intuitive and that authors don't even bother to prove, so I think there's something stupid that I'm not seeing. 
Let $M$ be a finitely generated module over $kG$, $k$ is a field, $G$ a group; and let $n$ be the minimal dimension over $k$ of a projective module that maps onto $M$; and let $P\to^\epsilon M$ be such a surjection, $P$ projective of dimension $n$. 
Then this is a projective cover of $M$. 
I really don't see how $\ker \epsilon$ can be proved to be superfluous, and I'm assuming there are some basic results about superfluous morphisms/submodules that I'm unaware of (I just learned about these notions).
I've tried the basic stuff like naively assume $N+\ker\epsilon = M$ and finding a projective module that maps onto $N$, but I can't  control the dimension well enough there.
So I'm looking for hints as to why this is true. 
One way of proving this is if we know that projective covers exist : it is really easy to see that if $M$ has a projective cover, then it has minimal dimension (simply because any other epimorphism from a projective lifts to an epi onto this projective cover); so an answer may very well just explain why a projective cover exists. 
I'm only interested in the case where $G$ is finite (I'm not sure whether it's true otherwise); and even if I don't think it changes anything, if it helps, I'm willing to assume that $G$ is a $p$-group, where $p= \mathrm{char}k$
(Let me just note that the title comes from the fact that in this situation, $\ker\epsilon$ is also called $\Omega (M)$)
 A: This definitely isn't true as stated if $G$ is not finite (since if $G$ is infinite the minimal dimension of $P$ might be infinite, but then "minimal dimension" does not prevent adding more junk).  So, let us assume $G$ is finite.  Then the category of finitely generated $kG$-modules is self-dual, by sending a finite-dimensional representation to its dual representation.  Since this category has injective envelopes, it follows by dualizing that it also has projective covers.  Explicitly, you can construct an projective cover of $M$ by taking an injective envelope $M^*
\to E$ of its dual and then dualizing to get $E^*\to M$ which is a projective cover.  (Note that $E$ will indeed be finite-dimensional, since $M^*\otimes kG$ is injective and finite-dimensional and there is a monomorphism $M^*\to M^*\otimes kG$ induced by the map $k\to kG$ sending $1$ to $\sum_{g\in G} g$.)
Alternatively, you can just imitate the proof of the existence of injective envelopes.  To find an injective envelope of a module $M$, you take a monomorphism $M\to I$ where $I$ is injective, and then take a maximal submodule $E$ of $I$ which contains $M$ as an essential submodule, and show that $E$ is a direct summand of $I$ and hence also injective by taking a maximal submodule of $I$ which is disjoint from $E$ and showing it is a direct complement of $E$.  This argument works in any abelian category, except that the maximal subobjects in question may not exist.  But if $I$ happens to be Noetherian, there is no issue and the maximal subobjects must exist.  So dually, in any abelian category with an epimorphism $P\to M$ where $P$ is projective and Artinian, we can find a quotient of $P$ which is a projective cover of $M$.  In the case of finitely-generated $kG$-modules, everything is Artinian and so this always works.
