Find a group with four elements in which every element is its own inverse 
*

*What is the procedure for solving this problem.
It seems like I can just say let $G$ be a group $\{1, a, b ,c\}$ and define $a * a^{-1} = b * b^{-1} = c * c^{-1} = 1$ ... but then what does $a*b$ or $b * c$ represent for example. I really have no idea for the procedure involved in solving the problem.

*How would I find a group with four elements in which not every element is its own inverse?

 A: *

*Take the multiplicative group $\{-1,1\}\times \{-1,1\}$ with pointwise multiplication.

*Take $\mathbb Z_4$, the equivalence classes modulo $4$.

A: Just to add something to 1.: There are several ways to create arbitrary big groups where every element is its own inverse.
For example the set of all bitstrings of length M with XOR as Group product has this property. That is why they are also called Boolean Groups.
(And they have to be commutative because ($ab=(ab)^{-1}=b^{-1}a^{-1}=ba$).)
You could also think of an alphabet with $n$ letters.
Then create a set that consists of all sequences of letters in alphabetic order where no letter appears more than once.
For example 'abcde' or 'dfgx' or just 'a' would be elements of the set.
Then define the empty sequence, ' ', as Identity and the Group Product shall be the ordered combination of two sequences from which all doubles are removed.
For example, the product of 'abc' and 'bdg' would be 'acdg'.
Then every element is its own inverse and it is quite different from XOR bitstrings because the sequences can all have different lengths.
The number of elements in the Group defined above are obtained as follows.
Having $n$ different letters, we can first build $1$ element (=sequence) which contains them all, then we can create $n$ sequences with $n-1$ letters by removing one of the letters.
Then we can create $n(n-1)$ sequences with $n-2$ letters but there will be doubles because it does not matter whether one first removes 'a' and then 'b' or the other way round. Thus there are
$\frac{n(n-1)}{2}=\frac{n!}{(n-2)!2!}= \begin{pmatrix}n\\2\end{pmatrix}$ different sequences with $n-2$ letters.
Iterating the process and taking the doubles out, there are
\begin{equation*}
    N = 1+n+\begin{pmatrix}n\\2\end{pmatrix}+\cdots+
        \begin{pmatrix}n\\n-1\end{pmatrix}+\begin{pmatrix}n\\n\end{pmatrix}
       = \sum_{i=0}^{n} \begin{pmatrix}n\\i\end{pmatrix}
\end{equation*}
elements in the set. Observe that the prelast contribution is due to $n$ different sequences with only one letter and the last contribution is the sequence without letters, ' ', which we defined as the Identity.
A: Your thought on 1 is correct, now you need to specify what $a*b$ is.  If you think of the group operation table, you must have one of each element in each row and column.  If the first element is the row of your table, you have already used $1$ and $a$.  You have also used $1$ and $b$ in the column, so the answer must be $c$.  As nothing before this has distinguished between $a, b, c$, we must have that the product of any two is the third.  This proves that if there is such a group, it must have the table $$\begin {array}{c c c c} 1 & a & b & c \\a&1&c&b\\b&c&1&a\\c&b&a&1 \end{array}$$  We have not proved that this is a group, with the only question being associativity.  I would take the question to promise that.
For 2, again let the elements be $1,a,b,c$.  One pair can be inverses, so there must be one that is its own inverse.  So we have $a*a=1=b*c=c*b$  Again we can fill in the operation table, starting from $a*b=c$ (it can't be anything else) getting $$\begin {array}{c c c c} 1 & a & b & c \\a&1&c&b\\b&c&a&1\\c&b&1&a \end{array}$$  As others have remarked, this is addition modulo $4$, with $(1,a,b,c)$ corresponding to $(0,2,1,3)$ though $b$ and $c$ can be reversed.
A: 1) An element $a$ is equal to its inverse, $a=a^{-1}$ iff $a^2=1$. Also if any element is its inverse then $ab=(ab)^{-1}=b^{-1}a^{-1}=ba$, so the group is abelian. 
Say the four elements of the group are $1, a, b, c$ then $ab=c$ and also it follows that  $bc=a, ca=b$.
An explicit example is (using addition mod 2) identity $(0,0), a=(1,0), b=(0,1), c=(1,1)$
2) If you want an example of a group of order 4 where some element is not its inverse use the integers mod 4; the element 2 is its own inverse but the inverse of 3 is 1.
A: Consider reflection, $\mu$, and 180 degree rotation, $\rho^2$, of a square.
It's easy to see $\mu^2 = \rho^4 = e$, and $(\mu \rho^2)^2 = e$.
Thus every element in $\langle \{\mu, \rho^2\} \rangle  = \{ e, \mu, \rho^2, \mu \rho^2 \}$ is its own inverse.
This is isomorphic to the Klein four-group, the direct group product $\langle \mu \rangle \times \langle \rho^2 \rangle$, the elementary abelian group $(\mathbb{Z} / 2\mathbb{Z})^2 = \{(0,0), (0,1), (1,0), (1,1)\}$ (with componentwise addition), and the dihedral group $D_2$ (replace $\rho^2$ with $\rho$).
Also it is easy to verify this set is a subgroup of $D_4$ and so is a group itself (in fact, normal to $D_4$).  
