Understanding the concept of Groups Currently, I am studying Math's C in grade 11 and we have started a unit known as Groups. I'm really struggling to understand the concepts of how all of this works, as the maths that we do is unlike anything that I have ever done before. 

These are the 'concepts' that we started off with, we then had to use these to determine whether or not equations were closed or associative as well as finding identities and inverses of operations. This is the point where the course really started to lose me. 

This is an example of the sort of questions we have been asked to do. My teacher does go through and explain the solutions to us, however; I barely understand why most things are done and how the final 'solution' actually proves anything.

The above solution was provided by my teacher, yet, I fail to see how either of the solutions answer the questions. Additionally, I don't understand where a majority of the variable came from e.g. a, b, 2x, 2y and 2z.
I just need some clarity on how this topic works, because the math seems so strange to me, and I just don't understand appropriate methods of solving the questions given to me.
 A: Groups are certainly an abstract thing to teach in high school. It's usually one of the first things taught about abstract mathematics, and it's not uncommon for them to leave people alienated. However, as one of my professors once told me, group theory is probably the most widely applicable branch of mathematics, so I guess it's important.
I suspect that one problem you're having is with motivation. You have this definition, with a checklist of four conditions that must be satisfied, but you don't know why people bother. What do these conditions actually form? What could you use these things for? Does anything interesting come out of them?
If I remember correctly, group theory was originally formulated based on groups of symmetries. For example, take a three dimensional cube. One easy way to represent the cube is via the eight points on the corners. Say we label the corners $1, 2, 3, 4, 5, 6, 7, 8$ like this:

A symmetry (in this case) is a way to change the orientation of the cube, in such a way that the cube "looks the same" at the end. For example, we could rotate the cube by taking $1$ to $2$, $2$ to $3$, $3$ to $4$, and $4$ to $1$. This would force $5$ to $6$, $6$ to $7$, $7$ to $8$, and $8$ to $5$.
Each symmetry will keep the cube looking the same, but takes the individual corners to potentially new corners. By keeping track of how the symmetry moves corners to corners, you can then figure out exactly how the entire cube moves.
Each symmetry "permutes" the set $V = \lbrace 1, 2, 3, 4, 5, 6, 7, 8 \rbrace$. That is, it forms a function $f : S \to S$ that is one-to-one and onto. That is,


*

*It can never take two different corners, and put them in the same corner (the cube would lose a corner!)

*Every corner must have a corner mapped to it by the function; no corner can be missed out.


Such symmetries form a group, under "composition". That is, we compose two symmetries $f$ and $g$ by first applying $g$, then applying $f$ (don't ask why it's in that order - it's convention). The resulting map is $f \circ g$, which is necessarily a symmetry, since the cube should "look the same" after both $f$ and $g$ happen.
For example, suppose $f$ was the rotation above, and $g$ rotates $2 \to 6, 6 \to 7, 7 \to 3, 3 \to 2$ and correspondingly, $1 \to 5 \to 8 \to 4 \to 1$. Then $f \circ g$ will take the point $1$ first to $5$ (under $g$), then the point $5$ to $6$ under $f$, so $f \circ g$ takes $1 \to 6$. You can work out how $f \circ g$ maps $2, 3, \ldots, 8$ as well if you wish, but that's how composition works.
The identity of the group of symmetries is the "identity" symmetry: the one that does nothing. That is, $1 \to 1$, $2 \to 2$, etc. And, for every symmetry, there's another symmetry that "undoes" it, which form the inverses. For example, $f$ being the rotation as above, we get $f^{-1}$ by rotating the cube 90 degree counter-clockwise. If you perform $f$ then $f^{-1}$ (or the other way around), in total, nothing happens to the cube, so $f \circ f^{-1}$ and $f^{-1} \circ f$ are the identity symmetry.
So, first question, can we jumble up the set $V$ of vertices any way we want with symmetries? No. For example, we couldn't map $1 \to 7$ and $5 \to 4$ at the same time. Why? Because $1$ and $5$ used to share an edge, but $4$ and $7$ don't. So, not every permutation of $S$ comes from a symmetry of the cube. Even if we respect edges, we can get into trouble; take a reflection like $1 \to 2 \to 1$, $5 \to 6 \to 5$, $8 \to 7 \to 8$, and $4 \to 3 \to 4$. We can write it down on paper, but there's no physical way to rotate the cube so that this becomes true.
That is, we get a group that is a little more interesting than just the group of permutations on $V$. Some people get really into these kinds of groups of symmetries. For example, chemists exploit various symmetry groups of molecules in order to help detect them, separate them from other molecules, and predict certain properties that the resulting chemicals will have. Knowing certain group theoretic properties can help you better understand these molecules.
It was symmetry groups that inspired the more general group definition, but groups are very interesting to a lot of people in their own right. We can look at so many common number systems/structures as groups in their own right: e.g. the real numbers, complex numbers, integers, rationals, symmetries, invertible matrices, sets under symmetric difference, vector spaces, modulo classes, vector transformations for systems of differential equations, and just a whole heap more. So many of these disparate structures can be united by a single theory; each single fact you learn about groups tells you something important about all of these structures.
Group theory has to be general and independent of all context, so that it can be used generally. It makes it seem dry at first, but it's a useful tool in many applications of mathematics. Groups are very powerful given the relative simplicity of the definition. But, you need to know how to apply the definition (check closure, associativity, identity, and inverses) before you can use any of the powerful theorems.
I hope this helps. I haven't really answered your question, but I do think motivation might be the first stumbling point here.
A: Maybe a little change in the notation will help you. Imagine a set of objects, you can name them whatever you want, say $S = \{A, B, \cdots\}$ What's really beautiful about this is that these objects can be anything (real numbers, integers, matrices, rotations, translations, ...) The trick here is that you need also one operation that takes two of these objects and transforms it into a third object that also belongs to the same set. 
To give you an example, imagine you are in a line, and the object $A$ means "move $a$ feet forward", similarly $B$. Now the operation: let's use a symbol $\smile$ (use whatever symbol you want), such that $A {\smile}B$ means, move $b$ and then move $a$. Intuitively there should be another object $C$ which is the result of operating $A$ and $B$, and is just the result of moving $A\smile B = a+b$.
Here are other two properties: 


*

*Staying at the same location (not moving) is also a possible element (existence of identity)

*If you walk forward $a$ feet, it is also possible to walk backward by the same amount, such that your final position is equivalent to not moving at all (existence of inverse)
Turns out that if the objects $S = \{A, B, \cdots \}$ together with the operation $\smile$ satisfy some axioms, you can build another object called a group. That satisfy


*

*For $A$ and $B$ in $C$, the element $A\smile B$ is in $S$

*For all $A$, $B$ and $C$ in S we have $A\smile (B \smile C)$ is the same as $(A \smile B)\smile C$

*There exists an element $I$ in $S$ such that for all $A$: $A\smile I = A$

*For all $A$ in $S$ there exists an element $\hat{A}$ such that $A\smile \hat{A} = I$
Now, the meaning of the objects and the operation can change from case to case. As in the example above there are plenty. Here's another one: $S = \{1, -1 \}$ with $\smile = \cdot$ (the multiplication symbol) (can you show that that is a group?)

EDIT
To go back to your example, imagine this case: $S = \mathbb{R}$ and $a\smile b = a + b -4$, the question is does $\{S,\smile\}$ form a group?. Let's try all the test 


*

*Closure: Imagine two numbers $a,b\in S = \mathbb{R}$, then $a\smile b = a + b - 4$, you know from school that if you add real numbers together, the result is a real number, so $a + b - 4$ is real, so $a\smile b$ is real, that is $a\smile b \in S$, and like that we just proved the first property

*Associativity: Consider now three numbers $a,b,c\in S$, and note that 
$(a\smile b)\smile b = (a + b - 4)\smile c = (a + b - 4) + c - 4 = a + b + c - 8$, similarly $a\smile(b\smile c) = a\smile(b + c - 4) = a + (b + c - 4) - 4 = a + b + c - 8$. So putting these two results together $(a \smile b)\smile = a \smile (b \smile )c$

*Identity: Is there a number $e \in S$ such that $a\smile e = a$, the answer is yes: $e = 4$: $a \smile 4 = a + 4 - 4 = a$, so identity of the group $\{S, \smile \}$ is $4$

*Inverse: Given $a$ can we find an element $b \in S$ such that $a\smile b = e$? And again the answer is yes, $b = 8-a$, indeed: $a\smile b = a + (8 -a) - 4 = 4 = e$. Usually this element $b$ is denoted as $a^{-1}$, so that $a\smile a^{-1} = e$
Since $\{S,\smile \}$ follow all these four axioms, then it is a group, you can call it masterj's group! Here's a bonus:


*$a\smile b = b \smile a$, groups that follow that additional property are called Abelian so masterj's group is Abelian!

A: I had to leave partway through this answer, I'll come back and finish it when I'm able.
The other answers give great explanation and motivation for the theory of groups, so I'll contribute with something more concrete.

When you're asked to see if something is a group, it's really asking four questions. In particular, you need to check the four following things:


*

*Closure: If $a$ and $b$ are things in your set, is it always the case that $a *b$ is in your set?

*Associativity: If $a,b,c$ are things in your set, is it always the case that $(a * b) * c = a * (b * c)$?

*Identity: Is there some $I$  in your set such that for any $a$ in the set, we have that $a * I = a = I * a$?

*Inverses: If $a$ is in the set, can you always find some other $a^{-1}$ in the set such that $a * a^{-1} = I$?


If the answer to all four of these questions is yes, then you have a group. If not, it's not a group.

Let's work through some examples. I'll do the three examples at the same time, so you can see the similarity in the arguments: these questions are really similar once you have some practice.
The three examples are: The even integers under addition, the even integers under multiplication and the operation $*$ you defined in the question.
Let's check the four things in turn. First, we need to check closure.
Example 1: Even integers under addition
Suppose $a$ and $b$ are arbitrary elements in the set. The set is just the set of even integers, so we know $a$ can be written as $a=2x$ for some integer $x$, and we know $b = 2y$ for some integer $y$. We'd like to show that $a+b$ is also an even number.
But $a+b = 2x+2y=2(x+y)$, so $a+b$ is twice some integer $x+y$, hence $a+b$ is an even number and is also in the set.
This is exactly what we needed! So, the set is closed under addition.
Example 2: Even integers under multiplication
Suppose $a$ and $b$ are arbitrary elements in the set. The set is just the set of even integers, so we know $a$ can be written as $a=2x$ for some integer $x$.
We'd like to show that $a \times b$ is also an even number. But $a \times b = 2 x b = 2(xb)$, so it is twice some integer. Hence $a \times b$ is an even number, and is also in the set. So, the set is closed under multiplication.
Example 3: Real numbers under $*$ 
Suppose $a$ and $b$ are arbitrary elements in the set, i.e. they are real numbers. We'd like to show that $a * b$ is also in the set, i.e. a real number. But $a* b = a + b - 4$, and so it is just the sum and difference of real numbers, so is clearly a real number. Hence, $a * b$ is in the set, and the set is closed under $*$.
Notice the similarity in all these arguments. We took two unknown things in the set and showed that the operation on them gave something else in the set. For more examples, try:


*

*Is the operation $a*b = \sqrt{|ab|}$ closed on the rational numbers?

*Is the operation $a*b = a-b$ closed on the integers?

*Is the operation $a*b = a-b$ closed on the positive integers?

*Is the operation $a*b = |a-b|$ closed on the positive integers?

*Is the operation $a*b = a \ln b$ closed on the real numbers?



Next, we need to check associativity.  
To understand the importance of the associative law, it's important to remember that an operation is a way of combining two elements; so if we want to combine three elements, we can do so in different ways. If we want to combine $a$, $b$, and $c$ without changing their order, we may either combine a with the result of combining $b$ and $c$, which produces $a * (b * c)$; or we may first combine $a$ with $b$, and then combine the result with $c$, producing $(a * b) * c$. The associative law asserts that these two possible ways of combining three elements (without changing their order) produce the same result.
The general strategy is the same: Take three arbitrary elements $a$,$b$,$c$ in the set, and check that $(a*b)*c = a*(b*c)$. 
Example 1: Let $a,b,c$ be even integers. Then they can be written $a=2x$,$b=2y$,$c=2z$ for integers $x,y,z$. 
$$(a+b)+c = (2x+2y)+2z = 2x+2y+2z$$
On the other hand,
$$a+(b+c) = 2x+(2y+2z) = 2x+2y+2z$$
So, $(a+b)+c = a+(b+c)$, exactly as we wanted.
Example 2: Let $a,b,c$ be even integers. Then they can be written $a=2x$,$b=2y$,$c=2z$ for integers $x,y,z$. 
$$(a\times b)\times c = (2x\times 2y)\times 2z = 4xy \times2z=8xyz$$
On the other hand,
$$a\times (b\times c) = 2x\times (2y\times 2z) = 2x\times 4yz = 8xyz$$
So, $(a\times b)\times c = a\times (b\times c)$, exactly as we wanted.
Example 3: Let $a,b,c$ be real numbers. Then
$$(a*b)*c = (a + b - 4)*c = (a+b-4)+c-4 = a+b-4+c-4=a+b+c-8$$
On the other hand,
$$a*(b*c) = a*(b+c-4) = a+(b+c-4)-4 = a+b+c-4-4=a+b+c-8$$
So, $(a*b)*c = a*(b*c)$, exactly as we wanted.

Now it's time for identity. It's always a good idea to do identity before inverses, because the rule for inverses uses the $I$ we found from identity. But, there's nothing wrong with doing identity before/after associativity, I just prefer to do it first because I learnt it that way.
Example 1:
We're looking for some $I$ such that for any $a$ in the set, $a + I = a$ and $I + a = a$. Well, we know that $x + 0 = x = 0 + x$ for integers, so perhaps $I=0$ would work... We can check! Let $a$ be in the set. Then it is even, so there is some integer $x$ with $a = 2x$. Then, $a + I = 2x + 0 = 2x = a$, and similarly $I + a = 0 + 2x = 2x = a$, so it works.
Example 2: 
This time, we're looking for some $I$ such that for any $a$ in the set, $a \times I = a$ and $I \times a = a$. Typically, we're used to the fact that $x \times 1 = x$ for integers, so that would be a nice idea. But, $1$ isn't an even integer, and we're only working in the set of even integers... So the obvious choice doesn't work. That doesn't mean that nothing works, however! We need a formal proof that nothing works instead. So, let's suppose (for the sake of contradiction) that for any $a$ in the set, $Ia = a$. Since it works for any $a$, it must in particular work for $2$. (I chose $2$ because it's the simplest even integer: I didn't pick $0$ because any $I$ works with $0$). Well, we now have $2I = 2$, and so we see that $I$ must equal $1$. But $1$ isn't in the set, so there is no possible identity.
(It's worth pointing out that your teacher's solution has a mistake in saying that there is an identity - as noticed in the original comments.)
So, we know that Example 2 doesn't form a group.
Example 3: 
We're looking for some $I$ such that $a * I = I * a = a$, in particular $a + I - 4 = a$. Looking at this equation, it seems like $I=4$ works, so let's prove it.
Let $a$ be a real number. We'd like to show $a * 4 = a$ and $4 * a = a$.
$a*4 = a + 4 - 4 = a$, as required. On the other hand, $4 * a = 4 + a - 4 = a$, which works too.
So, there is an identity and it is $4$.
A: Try using presentations:
If an group-algebraist writes 
$$\langle g\ |\ g^n=e\rangle,$$
that is specifying a group with one generator on which the relation implies that the group has 
$$\{g,g^2,...,g^{n-1},g^n=e\}$$
as its elements. It is realizable as the cyclic group $\Bbb Z_n$.
Other examples are
$$\langle a,b\ |\ a^2=e, b^3=e\rangle,$$
and
$$\langle a,b\ |\ a^2=e, b^3=e, ab=b^2a\rangle.$$
The 1st is realizable as the quotient $\frac{SL_2({\Bbb Z})}{H}$
were $H=\{1\!\!1,-1\!\!1\}$ but the 2nd is $S_3$ the symmetric group on a set with three elements. 
The general setting is
$$\langle g_1,g_2,...\ |\ R_1=1, R_2=1,...\rangle.$$
where the $g_i$ are called generators and the $R_j=1$ are conditions which govern the ways of reducing the operations among the $g_i$'s.
More examples are
$$\Bbb Z_2\oplus\Bbb Z_3=
\langle a,b\ |\ a^2=e,b^3=e, ab=ba\rangle.$$
$$\Bbb Z\oplus\Bbb Z=
\langle a,b\ |\ aba^{-1}b^{-1}=e\rangle.$$
$$\Bbb Z\rtimes\Bbb Z=
\langle a,b\ |\ abab^{-1}=e\rangle.$$
$$\Bbb Z*\Bbb Z_2=
\langle a,b\ |\ b^2=e\rangle.$$
Note: the symbol $1\!\!1$, above, is the identity matrix 
$\left(\begin{array}{cc}1&0\\0&1\end{array}\right)$. 
