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!


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.

  • $\begingroup$ 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

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 $\mathbb{Z}_7^\times$


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.

  • 1
    $\begingroup$ thanks for the fun fact, that should be helpful in the future ! $\endgroup$ – Justin Mar 10 '15 at 0:33

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 ?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.