# Does this set with this operation define a group?

Consider the set $A = \{a_1,a_2,...,a_n,e\}$. Here, $e$ will be our identity. Define the operation * such that $a_i*a_j = e$ for all $i,j\in\{1,2,...,n\}$ and $k*e = e*k = k$ for all $k\in A$. Is this a group?

I feel that it satisfies all three properties: associativity, existence of an identity, existence of an inverse for each element. However, clearly the inverses here are not unique... which goes against the easily derivable fact that inverses are unique in groups. What's going on intuitively here?

EDIT: Oh wow, it's not associative, thanks guys.

• It is non-associative if the $a_i$s are unique. – John Douma Oct 30 '15 at 21:35
• It seems to me we don't have associativity : $(a_1 * a_2) * a_3 = e * a_3 = a_3$ while $a_1 * (a_2 * a_3) = a_1 * e = a_1$. – Joel Cohen Oct 30 '15 at 21:38
• The equation $k*e=e*k=e$ defines an absorption element, not an identity element. (Contrast the numbers $0$ and $1$ with respect to multiplication.) Did you mean $k*e=e*k=\color{Red}{k}$? John and Joel's comments and Nitrogen's answer are based on the latter interpretation. If you intended the first, then your operation is associative but has no identity element. – whacka Oct 30 '15 at 21:38
• As you've defined it, it looks like you've set $a*b=e$ for all $a,b\in A$. This isn't a group either. – Milo Brandt Oct 30 '15 at 21:39

The problem here is associativity. Indeed, you have $$a_i=a_i*(a_j*a_k)\ne (a_i*a_j)*a_k=a_k$$ unless $a_i=a_k$ for every $i,k$, i.e., there is at most one non trivial element $a$. So $A=\{e,a\}$ which is of course isomorphic to $\Bbb{Z}/2$.
Edit: As whacka pointed out, I assumed that you meant $k*e=e*k=k$, otherwise it is obvious that $A=\{e\}$.
• The presence of an absorption element $e$ does not imply $A=\{e\}$. Indeed, the multiplication table originally given would be $(n\Bbb Z/n^2\Bbb Z,\times)$ under multiplication. (Of course, if we further assumed $A$ was a group, that would force $A=\{e\}$.) – whacka Oct 30 '15 at 21:50
• Of course, I assumed that he really wanted $A$ to be a group, so I gave the conditions for this to happen. – Nitrogen Oct 30 '15 at 21:51