Subgroups

[lclark]

Let G={(a,b)|a,b is an element of all real numbers, a does not equal 0} is a group where the set of all ordered pairs of real numbers the first component is not zero. The group under * is defined by (a,b)*(c,d)=(ac,bc+d). Show that H, where H={(a,b) is an element of G| b=0} (the second component is zero) is a subgroup of G. Meaning that it has an identity, an inverse and is closed.

 

[SlipEternal]

What help do you need?

 

[lclark]

I need help with the inverse

 

[SlipEternal]

Well, take what you start with.

Let \(\displaystyle (a_1,b_1)\in H\). We want \(\displaystyle (a_2,b_2)\in H\) such that \(\displaystyle (a_1,b_1)*(a_2,b_2) = e\) where \(\displaystyle e\) is the identity element you found before. You know that \(\displaystyle b_i=0\) for any \(\displaystyle (a_i,b_i)\in H\). So, \(\displaystyle (a_1,0)*(a_2,0) = (a_1a_2, 0)\). Therefore, it must be that \(\displaystyle a_2=a_1^{-1}\). That is it. That's the whole proof.

 

[lclark]

So since I found the identity to be (1,0). the inverse would be (-1,0)?

 

[SlipEternal]

No. You have a real number \(\displaystyle a_1\), you want to multiply it by another real number \(\displaystyle a_2\) and get \(\displaystyle 1\). As I wrote, \(\displaystyle a_2 = a_1^{-1}\). So, the inverse of \(\displaystyle (a_1,0)\) is \(\displaystyle (a_1^{-1},0)\). Now, check: \(\displaystyle (a_1,0)*(a_1^{-1},0) \stackrel{?}{=} (1,0)\)

 

[lclark]

Okay. The only way to get 1 would be to have the inverse equal (1,0), right. But can that be the inverse?

 

[SlipEternal]

Every element of the subgroup needs an inverse. The identity element is always its own inverse. So, given any element \(\displaystyle (a,0)\in H\), it has a distinct element \(\displaystyle (a,0)^{-1}\in H\) such that \(\displaystyle (a,0)*(a,0)^{-1} = (1,0)\). As I wrote, that inverse is going to be \(\displaystyle (a,0)^{-1} = (a^{-1},0)\)

 

[lclark]

I am sorry, I am not quite understanding. I understand that (1,0) will be its own inverse. However if a=2 then (2,0), then the inverse would be
(-2,0)? But I think what I am trying to ask is is there a single inverse for this subgroup, like (1,0).

 

[pka]

I understand that (1,0) will be its own inverse. However if a=2 then (2,0), then the inverse would be (-2,0)?

No the inverse of \(\displaystyle (2,0)\) is \(\displaystyle \left(\frac{1}{2},0\right).\)

 

[lclark]

Okay!, got it. and since it is all real numbers it would have an inverse. Just not a set in stone this one pair will work for all, correct?