What does Z_m*/<2> means?

Question

What is this notation? I think Z_m* is a group whose number is i \in Z_m s.t. GCD(i,m)=1 and <2> is a cyclic group

for example m=1023, the order of <2> is 10 since 1024=1 mod 1023 I cannot understand the slash / What is this?

Answer 1

The group $(\mathbb{Z}/m)^*$ has order $\phi(m)$, so $G=(\mathbb{Z}/1023)^*$ has order $600$. It is given by $(\mathbb{Z}/3)^*\times (\mathbb{Z}/11)^* \times (\mathbb{Z}/31)^*\cong C_2\times C_{10}\times C_{30}$. Determine the subgroup $U$ generated by $2$ in $G$. By Lagrange, the oder of $U$ is a divisor of $600$. Then determine the quotient group $G/U$ (the "slash").

Remark: I write $\mathbb{Z}/m$ instead of $\mathbb{Z}_m$, to avoid confusion with $p$-adic integers, for $m=p$.

Answer 2

There's actually a lot going on in your question. We start with $\mathbb{Z}_m$, which is also written as $\mathbb{Z}/m\mathbb{Z}$, $\mathbb{Z}/m$, and $\mathbb{Z}/(m)$. You can think of this as the set $\mathbb{Z}_m = \{0, 1, 2, \ldots, m - 1\}$ with addition and multiplication defined modulo $m$. As you've undoubtedly observed, multiplication behaves somewhat strangely. For example, in $\mathbb{Z}_{15}$, we have $9 \times 5 = 0$, meaning that two non-zero elements can multiply to zero! Elements like this are called zero-divisors. On the other hand, we also have $7 \times 13 = 1$, meaning that both $7$ and $13$ are invertible. A straightforward argument shows that zero-divisors don't have inverses. It's also easy to show that the product of two invertible elements is also invertible.

The notation $(\mathbb{Z}_m)^{\times}$ is used for the set of all (multiplicatively) invertible elements in $\mathbb{Z}_m$. A little work shows that $(\mathbb{Z}_{15})^{\times} = \{1, 2, 4, 7, 8, 11, 13, 14\}$. In fact, an element $k \in \mathbb{Z}_m$ is invertible precisely if ${\rm gcd}(k,m) = 1$; in other words, $k$ is invertible if it is relatively prime to $m$. (Exercise: Show that if ${\rm gcd}(k,m) > 1$, then $k$ is a zero-divisor.) But $(\mathbb{Z}_m)^{\times}$ is more than just a set, it's a group! We call it the group of units of $\mathbb{Z}_m$.

Oh, and since the Euler $\varphi$-function counts the number of positive integers that are relatively prime to (and less than) its input argument, we have: $|(\mathbb{Z}_m)^{\times}| = \varphi(m)$.

Two key facts about $\varphi$:

1. If ${\rm gcd}(a,b) = 1$, then $\varphi(ab) = \varphi(a)\varphi(b)$.

2. If $p$ is prime, then $\varphi(p) = p - 1$.

These facts imply, for example, that $|(\mathbb{Z}_{15})^{\times}| = 8$ and $|(\mathbb{Z}_{1023})^{\times}| = 600$.

In your example, the notation $\langle 2 \rangle$ denotes the subgroup of $(\mathbb{Z}_{1023})^{\times}$ generated (multiplicatively) by the element $2$. So: $\langle 2 \rangle = \{ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512 \}$.

The answer to your final question is a bit more involved, but $(\mathbb{Z}_{1023})^{\times}/\langle 2 \rangle$ is the quotient group formed from the cosets of $\langle 2 \rangle$. One such coset is $3\langle 2 \rangle = \{ 3, 6, 12, 24, 48, 96, 192, 384, 513, 768 \}$. Since $|\langle 2 \rangle| = 10$, the quotient group has $60$ elements (each coset has $10$ elements and the parent group has $600$ elements.)

So, the quotient group is an abelian group of order $60$. How many such groups are there? (Hint: Not many!)