# To which group is the $\mathbb Z_{20}^*$ isomorphic?

I have a question saying that to which group is $\mathbb{Z}_{20}^{*}$ is isomorphic, where $\mathbb{Z}_{20}^{*}$ is the set of the not zero divisors of $\mathbb{Z}_{20}$.

Here is what i think: $\mathbb{Z}_{20}^{*}=\{1,3,7,9,11,13,17,19\}$. It has 8 elements. Then it is isomorphic to $\mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2}$ or $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$ or $\mathbb{Z}_{8}$. But how can I decide which of them? And is this group a multiplicative group or an additive group? How can I know?

Thank you

-

Hints:

1. Is $1 + 3$ in $\Bbb Z^*_{20}$? On the other hand, one can show that it's closed under multiplication.
2. Is $\Bbb Z^*_{20}$ cyclic? What's the order of $3$? Answer both questions and one possibility will remain.

To elaborate on the second hint. We have: \begin{align} 3^1 &= 3 &\equiv 3 \pmod{20} \\ 3^2 &= 9 &\equiv 9 \pmod{20} \\ 3^3 &= 27 &\equiv 7 \pmod{20} \\ 3^4 &= 81 &\equiv 1 \pmod{20} \\ \end{align}

Therefore the order of $3$ is $4$. This eliminates the possibility $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. With similar computations, you can see that none of the elements of $\Bbb Z^*_{20}$ generates the group. Hence, it's not cyclic. One possibility remains and it's $\Bbb Z_4 \times \Bbb Z_2$.

To show that all elements of $\Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$ have order $1$ or $2$, consider $(a, b, c) \in \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2$. We have: $$(a, b, c)^n = (a^n, b^n, c^n)$$

Since each of $a$, $b$, $c$ is a member of $\Bbb Z_2$, it has order $1$ or $2$. This forces: $$(a, b, c)^2 = (a^2, b^2, c^2) = (1, 1, 1) = 1$$

-
I thought about looking at those but i cannot decide whether it is an additive or multiplicative group. How can i decide it? – bigO May 26 '13 at 22:27
@bigO It's multiplicative. I added an example showing how to calculate the order and elaborated on the hints. – Ayman Hourieh May 26 '13 at 22:34
Thank you, one more little question: how do we quickly decide that order of 3=4 eliminates z2xz2xz2? Why can't we have an element of order 4 in Z2xZ2xZ2? – bigO May 26 '13 at 22:37
@bigO Added some explanation. – Ayman Hourieh May 26 '13 at 22:40
Thank you very much – bigO May 26 '13 at 22:43

To hint your last question first, if $a,b$ are each relatively prime to $20$, then $ab$ is also relatively prime to $20$. $a+b$ need not be relatively prime to $20$.

To hint your first question, $\mathbb{Z}_8$ has an element of order 8, $\mathbb{Z}_2\times\mathbb{Z}_4$ has an element of order 4 but no element of order $8$, while in $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$ all elements but the identity have order 2.

-
well, i have some problems about checking the order of an element. For example, let us consider 3. What is its order? Should i look at it at modulo 20? – bigO May 26 '13 at 22:29
@bigO, here's how to calculate. $3^2=9$, $3^3=27\equiv 7$, $3^4=81\equiv 1$. Hence the order of $3$ is $4$, since this is the smallest exponent you need to return to the identity. – vadim123 May 26 '13 at 22:31
That is also what i thought but here i am confused about why we are looking at modulo 20. YEs we have Z20* but it has 8 elements. So i wonder why we look at modulo 20 but not modulo 8? – bigO May 26 '13 at 22:32
@bigO You look mod 20, since that's where these numbers are located. You've isolated 8 of them, but each and every one of them is still a representative of equivalence classes of numbers mod 20. – Arthur May 26 '13 at 22:39
@bigO You are getting confused. The group of units modulo 20 is a subgroup of the integers modulo 20. The elements you listed are still part of $\mathbb{Z}_{20}$ and should be treated accordingly. – Alex Wertheim May 26 '13 at 22:40

Hint By the Chinese Remainder Theorem $Z_{20}$ is isomorphic to $Z_{4}\times Z_5$. Also by the Chinese Remainder Theorem $Z_{20}^*$ is isomorphic to $Z_{4}^*\times Z_5^*$ [If you know rings, you can use directly the CRT for rings, otherwise you can prove the second statement from the standard CRT].

Since $5$ is prime $Z_5^*$ is a cyclic group with $4$ elements. $Z_{4}^*$ has 2 elements, and there are not too many groups with 2 elements...

-