Is it wrong to define $-1 \circ -1 = -1$ Problem:
Quaternions are a set of objects that are an extension of imaginary numbers except that there are three of them $i$, $j$ and $k$, with the relations
    \begin{align*}
    i^{2} = j^{2} = k^{2} = ijk = -1
\end{align*}
Construct the smallest group possible that contains all the quarternions.
My Solution:
Closure of the group requires that at least, $i,j,k$ and $-1$ to be the members of the group.
            Since $i^2 = i \circ i = -1$, $i$ can't be the identity of the group. Similarly $j$ and $k$ can't be identity of the group. That leaves $-1$ as the only candidate for the identity of the group. If we can satisfy other requirement of group, then $i,j,k$ and $-1$ will form a group with $-1$ as the identity.
If we define $-1 \circ -1 = -1$, which doesn't violate any of the given requirements, $-1$, works as the identity element.
Since $i^2 = i \circ i = -1$ and $-1$ is identity, $i$ by definition becomes the inverse of itself. Similarly $j$ and $k$ are inverses of themselves.
            So the group is 
            \begin{align*}
            G \left( \left\lbrace -1,i,j,k \right\rbrace, \circ \right)
        \end{align*}
Question: If all we are given is the statement on problem is it wrong to define $-1 \circ -1 = -1$ for this problem?
 A: As a teacher, I interpret the question as “What is the smallest subset of the ring of quaternions that contains $i$, $j$, $k$, and $-1$, and is a group under quaternionic multiplication?” I agree with the commenters that what is written is not the same thing.
Let us call this group $G$ for short.

Closure of the group requires that at least, $i,j,k$ and $-1$ to be the members of the group.

Yes.  But the closure can, and usually does, contain more than the elements used to define it.  

Since $i^2 = i \circ i = -1$, $i$ can't be the identity of the group. Similarly $j$ and $k$ can't be identity of the group. 

Indeed, identity elements are idempotents, so you are right that these non-idempotents can't be the identity.

That leaves $-1$ as the only candidate for the identity of the group. 

Here is where I think your logic breaks down. Nobody said that $i$, $j$, $k$, and $-1$ were the only elements of $G$.  Any product of elements of $G$ must also be in $G$.  So the group has to contain $i^3=-i$ and $i^4=1$ as well.
There are a few more quaternions that must be in $G$.  Can you find them?
