What is the difference between a Subgroup and a subset? I know hardly any Abstract algebra, just some things from youtube and wikipedia, but the notion of a subgroup being part of a larger group and a set being part of some group is indistinguishable to me. A nice simple answer would do fine. Thank you for your time
[Math] the difference between a Subgroup and a subset
abstract-algebraelementary-set-theorygroup-theoryterminology
Related Solutions
There's essentially no real difference between modules and representations. Think of them as two sides of the same coin.
Given a $\mathbb{K}G$-module $V$, you have a linear action of $G$ on a $\mathbb{K}$-vector space $V$. This in turn gives you a homomorphism from $G$ to $\mathrm{GL}(V)$ (invertible $\mathbb{K}$-linear endomorphisms). Such a homomorphism is a representation. And then this can be turned around. Given a representation, you get an associated module.
Specifically, let $V$ be a $\mathbb{K}G$-module and let $g,h \in G$, $v,w\in V$, and $c\in\mathbb{K}$. Give a name to the map: $v \mapsto g\cdot v$ say: $\varphi(g):V \to V$ (so $\varphi(g)(v)=g \cdot v$). Then $\varphi(g)(v+cw)$ $=g\cdot(v+cw)$ $=g\cdot v+cg\cdot w$ $=\varphi(g)(v)+c\varphi(g)(w)$. Thus $\varphi(g)$ is $\mathbb{K}$-linear. Then because $\varphi(1)$ is the identity map ($1 \cdot v=v$) and $\varphi(g^{-1})(\varphi(g)(v))=g^{-1}\cdot g\cdot v=(g^{-1}g\cdot v=1\cdot v=v$ etc. we get $\varphi(g)$ is an invertible linear map. Therefore: $\varphi:G \to \mathrm{GL}(V)$. Moreover, $\varphi(gh)=\varphi(g)\varphi(h)$ (easy to check) so $\varphi$ is a homomorphism (which we call a representation). Without going into the details, this all reverses.
So $\mathbb{K}G$-modules = representations of $G$ on $\mathbb{K}$-vector spaces.
If you've studied group actions, you've already seen this type of correspondence. Let $G$ act on $X$. Then the map $x \mapsto g \cdot x$ turns out to be a bijection on $X$. Thus if we define $\varphi(g)(x)=g\cdot x$ for all $x\in X$, then $\varphi(g) \in S(X)$ (permutations on $X$). Moreover, $\varphi(gh)=\varphi(g)\varphi(h)$ so $\varphi : G\to S(X)$ is a group homomorphism. We call such things permutation representations. And again this can be reversed. Given a permutation representation: $\varphi:G \to S(X)$, one can define a group action $g \cdot x \equiv \varphi(g)(x)$.
So $G$-action on $X$ = permutation representation of $G$ on $X$.
If you look into other branches of algebra, you'll see this kind of thing over and over again: Lie algebra modules = Lie algebra representations etc.
It's just different points of view. You can either think of "Algebra Thing" acting on "Thing" or a homomorphism from "Algebra Thing" to Maps from "Things to Things".
This is a tremendously common confusion to have, and in my experience, people are notoriously bad at explaining this concept. I'm sorry that you had to deal with people who were abrasive in addition to poor expositors.
In an arbitrary vector space, you cannot talk about components. They actually don't exist. Now, you can impose them on a finite-dimensional space by providing a bijective linear transformation from the arbitrary vector space to $F^n$, but then they're just that: an imposition, because any other bijective linear transformation will choose different would-be "components".
Components exist in $F^n$ because of the actual nature of the objects involved. So you don't need a basis, you can just look at an arbitrary object $(a,b,\dots,n)$, and find any of its components, because they're built into the object. This can be confusing because we also write coordinate vectors in this way, and when the basis is the standard basis, there is no difference between the components and the coordinates. However, in any other basis, there will be a difference.
(Edit: Val made an important point in the comments. I should have been more careful when I said there was "no difference". The fact is that coordinates and components are never conceptually the same, but I meant to say that in the standard basis case they will be numerically equal.)
Lacking a basis at all, you might want to say that $F^n$ still has coordinates implied by its components. But, in my opinion, this seems silly, since you cannot do the same in other spaces.
So the short answer is: Yes, there is a difference, because components are part of the objects.
As for your "collection of vectors" notion, they are basically the same. But it is easy to imagine a collection of vectors which is not a vector space: for example the circle in $\mathbb{R}^2$. This is definitely a collection, and the objects in it are definitely vectors, but it is not a vector space.
What I assume you meant by "collection" was what we might call a "meaningfully structured collection", and the meaningful structure is described precisely as an abelian group over which elements can be scaled by objects in a field. In that sense, your notion is correct, though a bit less transparent.
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Best Answer
Consider the group $\Bbb Z$ of integers under addition. The subset $S=\{1,2\}$ is not a subgroup. It isn’t a group at all: it isn’t closed under the operation (addition), since $2+2$ is not in $S$, and it doesn’t have an additive identity.
The set $E$ of all even integers, on the other hand, is a subset of $\Bbb Z$ that is a subgroup: it’s a group in its own right using the same operation, addition, as $\Bbb Z$.
If $G$ is any group, every subgroup of $G$ is by definition a subset of $G$, but as the example above shows, not every subset of $G$ need be a subgroup.