is this operating procedure an Abelian Group?

Question

I have to show if the following procedure gives a (Abelian) Group (G, *).

$G = \{ \textrm{true}, \textrm{false} \}$
$a*b := ( a \leftrightarrow b)$ (which means that $a$ is $\textrm{true}$ if and only if $b$ is $\textrm{true}$)

1.) Closure

For all $a,b \in G$, the result of the operation, $a * b$, is also in $G$. This is NOT given, since $a,b$ are not in $G$. However, the result of the operation $a*b$ is in $G$.

Well, do I have show "Associativity, ..." for this procedure, when $a,b$ are not in $G$? None of the group axioms are working.

Or am I misunderstanding this?

Thank you very much :)

Think of this as the group ${-1,1}$ where the operation is just usual multiplication — Prahlad Vaidyanathan, Nov 04 '13 at 09:36
Okay.. So a and b ARE elements of G? Let's say trfa = fa, trtr = tr and so on? — Vazrael, Nov 04 '13 at 10:01

Peter Taylor · Accepted Answer · 2013-11-04T14:49:27.970

2

Let's look at the operation. It seems to be XNOR, giving $$\begin{array}{ccc} * & \textrm{true} & \textrm{false} \\ \textrm{true} & \textrm{true} & \textrm{false} \\ \textrm{false} & \textrm{false} & \textrm{true} \\ \end{array}$$

By inspection of the table you have closure
What's the identity?
Does each element have an inverse?
How can you show that $(a \leftrightarrow b) \leftrightarrow c = a \leftrightarrow (b \leftrightarrow c)$? (I see a few options. The most straightforward and least elegant would be to draw up two 8-row truth tables)

edited Nov 04 '13 at 14:49

answered Nov 04 '13 at 10:59

Peter Taylor

13,425

Ok.. CLOSURE is ok. IDENTITY - I need an element e with ae=ea=a... Well.. mh.. ae=a means "TrueTrue=True" or "FalseTrue=False". In both cases e = true.. INVERSE: I need i with ai=ia=e. So truetrue=true; false*false=true. The inverse for a is a? – Vazrael Nov 04 '13 at 17:54
1

@K.L., yes, each element is self-inverse. – Peter Taylor Nov 04 '13 at 18:22

is this operating procedure an Abelian Group?

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