# Question on the Notation of an Abstract Algebra Question

The following is a question that I came across in a textbook I'm reviewing for self-study. The book is "Introduction to Abstract Algebra", 4th Edition, by W. Keith Nicholson. I have a question both about the notation. So here is the question in question:

1. In each case a binary operation * is given on a set M. Decide whether it is com-mutative or associative, whether a unity exists, and find the units (if there is a unity).

$$M = P \times Q$$, where $$P$$ and $$Q$$ are sets with $$|P| \ge 2$$ and $$|Q| \ge 2$$; (p,q)*(p',q') = (p,q')

...and my question:

What exactly is the difference between p and p'? I'm reading it as $$p,p' \in P$$, but $$p \neq p'$$. However, if that were the case then wouldn't $$P$$ have to contain the multiplicative unity, and how would they do that if both $$|P|$$ and $$|Q|$$ are $$\ge 2$$?

Gah. I am much confuse!

• The usage here is the same as saying $(p,q)*(r,s)=(p,s),$ and it is standard convention that different letters can correspond to the same element (i.e. $p=r$ is possible) UNLESS it is specifically stated (or can be deduced from something else) that $p \neq r$ must hold. – Dave L. Renfro May 24 '16 at 20:40

$p$ and $p'$ are both arbitrary elements of $P$. They are possibly equal, but need not be. Furthermore, $P$ need not have the concept of multiplication and therefore doesn't need to have the concept of a multiplicative identity. Multiplicative identities only make sense if we have a multiplication to talk about.

For example: $P=\{red,blue,yellow\}$ and $Q=\{dog,cat\}$

What is $dog$ times $cat$? It doesn't make sense, but we don't care! The operation is defined on $M=P\times Q$, not on $P$ and not on $Q$.

The operation $*$ in this case is defined as $(p,q)*(p',q')=(p,q')$

That is to say, given your first tuple $(p,q)$, and your second tuple $(p',q')$ the result of the operation is a tuple where the first entry is the same as the first tuple's and the second entry is the same as the second tuple's.

Following the example I started:

$(red,dog)*(blue,dog) = (red,dog)$

$(red,dog)*(blue,cat) = (red,cat)$

$(blue,dog)*(red,dog)=(blue,dog)$

Of course, we don't know exactly what $P$ and $Q$ look like in general, the above was just an example of how they might look. But, we do know that $P$ and $Q$ each have at least two elements.

Suppose then that $P=\{p_1,p_2,\text{possibly other stuff too}\}$ and $Q=\{q_1,q_2,\text{possibly other stuff too}\}$ where $p_1\neq p_2$. Maybe $p_1$ is a color, maybe it is a type of animal, maybe it is an integer. We don't know and we don't really care. Same for $q_1,q_2$.

Does it matter what order things are "multiplied"? That is, is $(p,q)*(p',q')=(p',q')*(p,q)$ in all cases?

look at my example with color,animal more closely. Try to generalize this for arbitrary $P,Q$

Does $((p,q)*(p',q'))*(p'',q'') = (p,q)*((p',q')*(p'',q''))$ for all $p,p',p'',q,q',q''$?

Try to see for yourself, what will wind up being the first entry of the resulting tuple? The second entry of the resulting tuple? Did it matter what order we did the operations in?

Is there an element $(1_p,1_q)$ such that $(p,q)*(1_p,1_q) = (p,q)=(1_p,1_q)*(p,q)$ for all $(p,q)$?

If it worked for $(p_1,q_1)$ does it also work for $(p_2,q_2)$?