Question

Find an isomorphism from the additive group Z6 to the multiplicative group G = {[a] is in Z7|[a] does not equal 0}. Click for picture.

Answer 1

\[Z _{6}, G={[a] \in Z _{7} | [a] \neq 0}\]

Answer 2

Is \(G\) just the group of all positive integers that are not multiples of \(7\)?

Answer 3

Thats the example from the book. I am not sure of anything else.

Answer 4

How about \(\phi:\mathbb{Z}_6^+\to\mathbb{Z}_7^\times\) s.t \(\phi\left([x]_7\right)=[2]^x_7\)?

Answer 5

$$\phi(x+y)=[2]_7^{x+y}=[2]_7^x\cdot[2]_7^y=\phi(x)\cdot\phi(y)$$

Answer 6

Though I'm not sure if it's bijective...

Answer 7

I do not understand any of that. How I have is it is how it is written.

Answer 8

Do you know what a group isomorphism is? It's a bijective homomorphism between two groups -- i.e. it shows they have the same underlying structure.

Answer 9

Yes

Answer 10

So then we're looking for an isomorphism between addition on some group of integers with multiplication on another group of integers; the obvious operation that comes to mind is exponentiation, since \(e^x\cdot e^y=e^{x+y}\).

Answer 11

ok

Answer 12

Now we just consider something like \([3]_7^x,\forall x\in G\):$$\phi([0]_6)=[3]_7^0=[1]_7\\\phi([1]_6)=[3]_7^1=[3]_7\\\phi([2]_6)=[3]_7^2=[2]_7\\\phi([3]_6)=[3]_7^3=[6]_7\\\phi([4]_6)=[3]_7^4=[4]_7\\\phi([5]_6)=[3]_7^5=[5]_7$$ so it is indeed a bijection!

\(\phi:\mathbb{Z}_6^+\to G\) is bijective as demonstrated above; every \([x]_6\in\mathbb{Z}_6^+\) has a corresponding unique \(\phi([x]_6)\in G\), and every \([x]_7\in G\) has a corresponding unique \(\phi^{-1}([x]_7)\in\mathbb{Z}_6^+\)

Answer 13

Okay so that would prove that this a part of the group?

Answer 14

it's clearly a homomorphism, too, since $$\phi([x]_6+[y]_6)=\phi([x+y]_6)=[3]_7^{x+y}=[3]_7^x\cdot[3]_7^y=\phi([x]_6)\cdot\phi([y]_6)$$

Answer 15

$$\phi([x]_6)=[5]_7^x$$ works as well.