1
$\begingroup$

Prove that that the following is or is not a group.

(a) The set S = $\mathbb{R}$ \ {0} with operation defined by a * b = 2ab for all a and b in S. (On the right side of the equation, the operations are the usual addition and multiplication in $\mathbb{R}$.)

(b) The set S = $\mathbb{R}$ with operation defined by a * b = 2(a + b) for all a and b in S.

I do not have the slightest clue how to begin. I know that to be a group the binary operation must be associative, the group must have an identity, and every element in the group must have an inverse.

The trouble I am having is identifying what the identity is in a) and starting the proof. I am also confused about part a) in that I do not know what equation is being referred to in the parenthesis. Please help?

$\endgroup$
  • $\begingroup$ First, you should recall how a group is defined. $\endgroup$ – quid Jan 17 '15 at 22:04
  • 1
    $\begingroup$ I know that to be a group the binary operation must be associative, the group must have an identity, and every element in the group must have an inverse. The trouble I am having is identifying what the identity is and starting the proof. $\endgroup$ – Karina Pena Jan 17 '15 at 22:09
1
$\begingroup$

The identity in (a) will be $1/2$: if $a * b = a$, then $2ab = a$, $a(2b-1) = 0$ and therefore we are forced to set $b = 1/2$ (because $*$ is commutative).

In (b) notice that the operation is simply the usual addition multiplied by two and $(\mathbb R, +)$ is a group.

$\endgroup$
1
$\begingroup$

I'll prove (a)-the proof of (b) is very similar. This is where you need to learn the importance of beginning with a clear understanding of the definition. The definition of the group is as follows: Def: Let (G,*) be an ordered pair where G is a nonempty set and let :G x G ----> G is a binary operation on G such that: (i) (Closure Axiom) For every a,b in G, ab is in G; (ii) (Associativity) For every a,b,c in G, (a*b)c= a(bc). (iii) (Identity) There exists a unique element e in G such that for every x in G, ex =xe =x; (iv) (Inverse) For every x in G, there exists a unique element y such xy=y*x =e.

Then (G,*) is a group. We usually just refer to the set G when referring to a specific group. This means to prove G is a group, we have to test the axioms.

So now let's consider the first problem. G = S = $\mathbb{R}$ \ {0} with operation defined by a * b = 2ab for all a and b in G. Consider first the closure axiom. Let a and b be in G. Then a * b = 2ab. Since a and b are nonzero real numbers and 2 is a positive real number, 2ab is a nonzero real number and therefore 2ab is in G and closure holds. Now we need to see if there is an identity element in G.Let a be in G. How do we find an element that leaves a unchanged in G under the defined operation? Let ae = 2ae =a. Since all the elements here are nonzero real numbers with the usual addition and multiplication, we can simply solve for e in the usual manner. So e = 1/2 in G and this is the identity under this operation. We now use a similar operation to find the inverse for a given element x. xy = 2xy= 1/2 -----> y =1/4 x. Again, since x and y are nonzero real numbers, y is also in G. So the inverse exists in G for every x. Lastly, we check associativity. Let x,y,z be in G. Then (x*y)z = 2xyz = 4xyz = 2* x * 2yz = x*(y*z). So associativity holds and G is indeed a group!

The second problem is proved the exact same way. Remember-the critical thing is being clear what the definition of the structure you're trying to establish is.

$\endgroup$
1
$\begingroup$

For (a) note that $f\colon \mathbb R\setminus\{0\}\to \mathbb R\setminus\{0\}, x\mapsto 2x$ turns out to be a bijection compatible with the operations, i.e., $f(a*b)=f(a)\cdot f(b)$, hence $(\mathbb R\setminus\{0\},*)$ is isomorphic to $(\mathbb R\setminus\{0\},\cdot)$, hence is also a group.

For (b) note that there cannot be a neutral element: If $a*e=a$ for all $a$, then we need $2(a+e)=a$, i.e., $a=-2e$ for all $a$, which is absurd.

$\endgroup$
  • $\begingroup$ Very clever,but might be a bit sophisticated for the OP,who strikes me as a rank beginner. $\endgroup$ – Mathemagician1234 Jan 17 '15 at 22:57
  • $\begingroup$ @Mathemagician1234 Admitted, but also for beginners it can be valuable to learn that otherwise lengthy computations especially for associativity and/or distributivity can be trivialised by exhibiting an isomorphism-to-be as here (alright, the computations are not that complicated here). - Also, I've seen solutions like this submitted to problems like this; and as these problems are only posed to beginners, there are beginners finding such solutions. $\endgroup$ – Hagen von Eitzen Jan 18 '15 at 12:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.