Abstract Algebra a big helping hand wanted :(

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Rebecca123

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Consider the binary operation ⊠ : Z6 x Z6 => Z6 given by
a ⊠ b := ab + 5(a + b) + 2

For each element in Z6, find its inverse with respect to ⊠ or prove that it doesn't exist.

You guys, not sure how to do this one at all.
What are your thoughts?:confused:
 
Rebecca123 said:
Consider the binary operation ⊠ : Z6 x Z6 => Z6 given by
a ⊠ b := ab + 5(a + b) + 2

For each element in Z6, find its inverse with respect to ⊠ or prove that it doesn't exist.

You guys, not sure how to do this one at all.
What are your thoughts?:confused:
Click to expand...

First, to get you started, what does the inverse mean? Given a in Z6, there exists an element b in Z6 such that a ⊠ b = b ⊠ a = Identity Element

So looks like you should find the Identity Element first.
 
Ishuda told you that the definition of "inverse" is that the inverse of element a is element b such that a ⊠ b= identity.

Here a ⊠ b= ab+ 5(a+ b)+ 2 (mod 6). So for each a in the set (there are only 6) its inverse is the element, b, satisfying ab+ 5(a+ b)+ 2= identity. For example, to find the inverse of "3" you want to find b such that 3b+ 5(3+ b)+ 2= identity (mod 6). You know what "identity" is so that is a simple linear equation.

At worst, you could calculate 3 ⊠ b for every b in the set (again there are only 6) and see which, if any, give the identity.
 
So what kind of an answer should I be coming out with?
Will I have to find the inverse 6 times because its Z6?:confused:
Thanks for you help by the way ;)
 
Rebecca123 said:
So what kind of an answer should I be coming out with?
Will I have to find the inverse 6 times because its Z6?:confused:
Thanks for you help by the way ;)
Click to expand...

According to the way the problem is stated, you will have to find the inverse for all six elements or, if an element doesn't have an inverse, prove that it doesn't.
 
Off the top of your head do you think an inverse exists for all?
Thanks for the help
 
Rebecca123 said:
Off the top of your head do you think an inverse exists for all?
Thanks for the help
Click to expand...
I haven't worked it out for all but just because of the way the problem is worded, I suspect there isn't an inverse for all.
 
I really don't know if this is right but I got 4 for the identity element in Z6.
Made out a few equations for the identity and arrived at (e-4)(a+1/2)=0
e-4=0 so e=4 identity element for Z6:confused:
 
Rebecca123 said:
I really don't know if this is right but I got 4 for the identity element in Z6.
Made out a few equations for the identity and arrived at (e-4)(a+1/2)=0
e-4=0 so e=4 identity element for Z6:confused:
Click to expand...

The identity element, I, for this set must satisfy
I ⊠ b = I b+ 5(I+ b) + 2 = (5+I) b + 5I + 2 = b mod(6)
for every b in the set.
4 ⊠ b = 9b + 22= 3b + 4 mod(6)
So, since 3b + 4 is not generally equal to b mod(6), 4 isn't the identity element for this set.
 
The identity, I, must have the property that a⊠I= aI+ 5(a+ I)+ 2= a for all a in the set. If you cannot solve that equation for I, with general a, try it with specific values:

If a= 1 that becomes 1⊠I= I+ 5(1+ I)+ 2= 6I+ 7= 1 so that 6I= -6 or I= -1. "mod 6" that is I=6- 1= 5.

If a= 2 that becomes 2⊠I= 2I+ 5(2+ I)+ 2= 7I+ 12= 2 so that 7I= (6+ 1)I= -10= -2(6)+ 2 or I= 2 (mod 6).

Those are NOT the same so it looks like there is NO such "I".
 
Ok I see, thank you so much.
Does this mean that an inverse doesn't exist either? How would I go about proving this?
Thank you to you all for your help, really appreciate it.
 
Rebecca123 said:
Ok I see, thank you so much.
Does this mean that an inverse doesn't exist either? How would I go about proving this?
Thank you to you all for your help, really appreciate it.
Click to expand...

Let a = 2 and consider 2⊠b
2⊠b = 2b+ 5(2+ b)+ 2= 7 b + 12 = b + 6(b+2) = b mod(6)
for all b in Z6, so 2 is the identity under ⊠

Let a = 1 and consider 1⊠b
1⊠b = 1b+ 5(1+ b)+ 2= 6 b + 7 = 1 + 6(b+1) = 1 mod(6)
for all b in Z6, so 1 has no inverse under ⊠

Go through the rest of the numbers in Z6 and see what you get.
 
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