# Prove that the following set is a group

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?

• First, you should recall how a group is defined. – quid Jan 17 '15 at 22:04
• 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. – Karina Pena Jan 17 '15 at 22:09

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.

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.

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.

• Very clever,but might be a bit sophisticated for the OP,who strikes me as a rank beginner. – Mathemagician1234 Jan 17 '15 at 22:57
• @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. – Hagen von Eitzen Jan 18 '15 at 12:45