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Why is the group $Z_6$ under addition isomorphic to the group $Z_7^*$ under multiplication?

So I'm trying to answer this question:

Q. Which of the set are isomorphic to each other?

$S_3$, $Z_6$, $Z_3 \times Z_2$ and $Z_7^*$

Now I know that $S_3$ is not because its not abelian while $Z_6$, $Z_3 \times Z_2$ are since my professor said this:

The group $Z_n$ under addition is an abelian group which means $Z_6$ is abelian.

Now another idea is: The direct product of two cyclic groups $Z_m$ and $Z_n$ is isomorphic to ($Z_{mn}$,+) iff m and n are relatively prime.

For $Z_3 \times Z_2$, 3 and 2 are relatively prime which means $Z_6$ and $Z_3 \times Z_2$ are isomorphic.

Now, whats throwing me off is $Z_7^*$. I know $Z_7^*$ is composed of the elements {1,2,3,4,5,6}. So i've researched a bit online, and $Z_7^*$ has an order of 6 which I'm not understanding.

Also, it was mentioned that $Z_7^*$ is abelian. Therefore, are all groups $Z_n^*$ under multiplication an abelian group? and why does $Z_7^*$ have an order of 6? Also, I've seen people say $Z_7^*$ is cyclic and has a generator such as 3, I'm not understanding that concept or in other words, how is 3 a generator? because if i do $3^6$=729 and that is not divisible by 7.

Or if anyone has a different way to do this problem, that would help as well. I'm open to any ideas. thanks!

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Let's look at the structure of $(\Bbb Z_7)^{\times}$ in some more detail.

$[1]$ isn't very interesting, it's clearly the (multiplicative) identity, though.

So now let's look at $[2]$, specifically, its powers.

$[2]^2 = [2]\cdot [2] = [4]$. This is...boring.

$[2]^3 = [2]\cdot[2]\cdot[2] = [8] = [1]$ (because $8 = 1 + 7$, so $8 \equiv 1$ (mod $7$)).

This tells us $[2]$ has order $3$.

OK, so now let's look at $[3]$.

$[3]^2 = [9] = [2]$. Thus $[3]^6 = ([3]^2)^3 = [2]^3 = [1]$. So the order of $[3]$ divides $6$ (this is, of course, self-evident by Lagrange).

As we saw above $[3]^2 = [2] \neq [1]$, so the order of $[3]$ is NOT $2$.

$[3]^3 = [27] = [6] \neq [1]$, so the order of $[3]$ is NOT $3$.

This means the order of $[3]$ must be $6$, so it generates the entire group:

$[3]^1 = [3]$

$[3]^2 = [2]$

$[3]^3 = [6]$

$[3]^4 = [4]$ ($81 = 4 + 7\ast 11$)

$[3]^5 = [5]$ ($243 = 5 + 7\ast 34$)

$[3]^6 = [1]$

Therefore, an isomorphism between $(\Bbb Z_6,+)$ and $((\Bbb Z_7)^{\times},\ast)$ is:

$[k]_6 \mapsto [3^k]_7$ (where the subscripts/brackets mean "equivalence class modulo").

Note $[3]^{-1} = [5]$, so we could have used $[5]$ as a generator instead.

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  • $\begingroup$ why did you take an elemento or the same order as the group and suddenly said that this defined an isomorphism between the 2 groups? Is there a theorem about it? Could you help me? $\endgroup$ – Guerlando OCs Jul 11 '15 at 19:40
  • $\begingroup$ The order of an element is the same as the order of the cyclic group it generates. $\endgroup$ – David Wheeler Jul 11 '15 at 22:02
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1 plays the same role in multiplication as 0 does in addition: $1\times a=a,0+a=a$. They are called the identity. (0 is the additive identity in $\mathbb{Z}_6$, 1 is the multiplicative identity in $\mathbb{Z}_7^\times$).
$3^6=729\equiv1\pmod7$, so six threes give the identity. Check that no smaller power of 3 is the identity, so 3 generates the whole group $\mathbb{Z}_7^\times$

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What's throwing you off is the definition of $Z_{7}^{*}$={${[n]_{7}: gcd(n,7)=1}$} and is a group under multiplication not addition. And yes all $Z_{n}^{*}$ are abelian since the multiplication of integers is abelian. And To show that it's cyclic with a generator of 3, just take all powers of 3(mod7) between 1 and 7 and you'll see that you get every element of the group. Now the group has order 6 because there are six elements in the group. And by a well know theorem, any finite cyclic group of order n must be isomorphic to $Z_{n}$. Also a fun fact, every element besides the identity is a generator in $Z_{p}$ where p is prime. Once you learn the sylow theorems, you'll see why.

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    $\begingroup$ thanks for the fun fact, that should be helpful in the future ! $\endgroup$ – Justin Mar 10 '15 at 0:33
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Because $(\mathbf{Z}/7\mathbf{Z})^{\times}$ is a cyclic group of order $6$, and any cyclic group of order $n$ is isomorphic to $\mathbf{Z}/n\mathbf{Z}$, by mapping any of its generators to $1$.

Now, why is $(\mathbf{Z}/7\mathbf{Z})^{\times}$ a cyclic group of order $6$ ? You have a conceptual reason : because it is the multiplicative group of the field $\mathbf{F}_7 = \mathbf{Z}/7\mathbf{Z}$. You have a down to earth reason : find a element $x$ in $(\mathbf{Z}/7\mathbf{Z})^{\times}$ such that $\{x^n\;|\;n\in\{0,\ldots,6\}\}$ is equal to $(\mathbf{Z}/7\mathbf{Z})^{\times}$. Do you see how to proceed ? Hint : what about $x = 3$ ? If it works, do you see why ?

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