group theory question

Question

Decide which of the following binary structures are groups. For each, if the binary structure isn't a group, prove that. If the binary structure is a group, prove that

a) (1/2 Z,+) where1/2 Z ={n/2 |𝑛 ∈ 𝑍}

b) (𝑅 ≥ 0,∗) 𝑤ℎ𝑒𝑟𝑒 𝑥 ∗ 𝑦 = 𝑚𝑎𝑥{𝑥, 𝑦}

c) (𝑅,∗) 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑎 ∗ 𝑏 = 𝑎 + 𝑏 + 𝑎𝑏 ∀ 𝑎, 𝑏 ∈ 𝑅

d) (𝑅⁺,∗) 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑎 ∗ 𝑏 = √𝑎𝑏 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎, 𝑏 ∈ 𝑅⁺

Answer 1

In order to determine if these are groups you must determine if it satisfies the following 3 properties or not:

1. Associative.
2. There is an identity element.
3. Each element has an inverse.

Part a) is a group so you must show that it satisfies all the properties above.

First we show it is associative:

Let x,y,z be arbitrary elements of 1/2Z.

Then by definition of 1/2Z there exists m,n,k in Z such that

x=m/2, y=n/2, z=k/2

Then (x+y)+z = (m/2+n/2)+k/2 by substitution

= ((m+n)+k)/2 by algebra

= (m+(n+k))/2 because + on integers is associative

= m/2+(n/2+k/2) by algebra

= x+(y+z)

Proving it is associative.

Next we show it has an identity element.

Notice that 0=0/2 is an element of 1/2Z, we will prove this is the identity element.

Let x in 1/2Z be arbitrary/

Then by definition there exists an n in Z such that x=n/2.

Now 0+x=0+n/2=n/2=x and x+0=n/2+0=n/2=x.

This proves 0 is the identity element.

Finally we need to show each element has an inverse.

Let x in 1/2Z be arbitrary/

Then by definition there exists an n in Z such that x=n/2.

Now notice that -n/2 is in 1/2Z because -n in Z.

Moreover x+-n/2 = n/2-n/2 = 0 and -n/2+x=-n/2+n/2=0.

Thus -n/2 is an inverse of x and because x was arbitrary we have proven every element has an inverse.

Thus we have shown it satisfies all the properties of a group and hence is a group.

For part b) notice not every element has an inverse.

Notice that for all x in R 0*x=max{0,x}=x and x*0=max{x,0}=x so that 0 is the identity element.

Now 2 in R but for any y in R, 2*y = max{2,y} which is greater than or equal to 2. Mainly no matter what element you choose you cannot have 2*y=0. Thus y does not have an inverse so it is not a group.

For part c) it also does not have an inverse for every element.

To see this notice a*0 = a+0+a0 = a and 0*a = 0+a+0a = a so 0 is the identity element.

Now -1 in R, but does not have an inverse because if it did say b was its inverse we would have:

-1*b = 0

But using the definition of * this means

-1+b+(-1)b = 0

-1 = 0, which is impossible.

Thus -1 does not have an inverse so it is not a group.

Finally for part d) it does not have an identity element. To see this assume it had an identity element call it x. Then it is the identity element for every element in R+ so in particular we have:

x*2=2 and x*3=3

Using the definition of * we have that

sqrt(2x)=2 and sqrt(3x)=3

Solving both equations for x yields

x=2 and x=3

But x is a single element and cannot be 2 and 3, thus we have a contradiction and no identity element exists.

Thus we have show it is not a group.

I hope this solutions helps! If you need any more help please feel free to contact me!