$\mathbb{Z}_{11}$ = {0,1,2,3,4,5,6,7,8,9,10}

+ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|---|

0 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 0 |

2 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 0 | 1 |

3 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 0 | 1 | 2 |

4 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 0 | 1 | 2 | 3 |

5 | 5 | 6 | 7 | 8 | 9 | 10 | 0 | 1 | 2 | 3 | 4 |

6 | 6 | 7 | 8 | 9 | 10 | 0 | 1 | 2 | 3 | 4 | 5 |

7 | 7 | 8 | 9 | 10 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |

8 | 8 | 9 | 10 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |

9 | 9 | 10 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |

10 | 10 | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

**Is $\mathbb{Z}_{11}$ a group?**

*Claim:* The set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with addition modulo 11, $\langle\mathbb{Z}_{11}, +_{11}\rangle$, is a group.

*Proof:*

$\mathcal{G}$1: First, let a +_{11} (b +_{11} c) = x and (a +_{11} b) +_{11} c = y in $\mathbb{Z}_{11}$. Then, let a + (b + c) = m and (a + b) + c = n in $\Bbb{Z}$. Since we know $\Bbb{Z}$ is associative under addition, m = n. By the division algorithm, we let m = 11q + r_{1} and n = 11q + r_{2} where 0 < r <11. Since m and n are equal, we know that r_{1} and r_{2} must be equal. Then, we say x = r_{1} and y = r_{2} because r_{1} and r_{2} are the remainders that represent the partitions x and y. Thus, x = y and associativity under addition holds for $\mathbb{Z}_{11}$.

$\mathcal{G}$2: Let e = 0 such that 0 + a = a + 0 = a.

$\mathcal{G}$3: Let a' = 11 - a. Then, 11 = a' + a and since 11 is not in the set (see above), a + a' = 0, since 0 +_{11} 11 = 0. Then, since 0 is the identity element, we know that a' must be the inverse of a. QED

**What is the order of $\mathbb{Z}_{11}$?**

The order of a group itself is the number of elements in the group and is denoted $|G|$. In other words, the order of a group is the cardinality of the set that belongs to the group. In our case, the order of $\mathbb{Z}_{11}$ is 11. We can look at the set $\lbrace 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\rbrace$ and easily see that the cardinality is 11 simply by counting.

**What is the order of the elements of $\mathbb{Z}_{11}$?**

we recall the cyclic group $H = \lbrace a^n | n$ $\epsilon$ $\mathbb{Z}\rbrace$ that is a subgroup of $G$. If this cyclic subgroup $\langle a \rangle$ is finite, then the order of a is the order $|\langle a \rangle|$ of the cyclic subgroup. In other words, if $a \epsilon$ $G$ is of finite order m, then m is the smallest positive integer such that $a^m = e$. Otherwise, we say that $a$ is of infinite order. In our case, we have already shown that each generator of $\mathbb{Z}_{11}$ creates a finite group. We also know that $e = 0$ for our group, so we must fulfill the equation $a^m = 0$. Since $0 \leq a \leq 10$, we know that m cannot easily be solved for. In order for $a^m$ to approach 0, m would have to be a very large, negative number and the definition states that m must be positive. Thus, the order of each element of $\mathbb{Z}_{11}$ is infinite.

**Is $\mathbb{Z}_{11}$ abelian?**

Theorem 6.1 in the Fraleigh book states that every cyclic group is abelian. The proof is relatively simple since the integers are commutative under addition. Basically, if we have two elements from our group a and b, such that $a = c^x$ and $b = c^y$, then $ab = c^xc^y = c^{(x+y)} = c^{(y+x)} = c^yc^x = ba$. We know that we can add and commute the x and y when they are exponents because here they are integers and we know that integers are commutative. We have already shown that $\mathbb{Z}_{11}$ is cyclic, and so it follows that $\mathbb{Z}_{11}$ is also abelian.

**What are the generators of $\mathbb{Z}_{11}$?**

We found, in class, that, in general, anything that has a common divisor with n (for the group $\langle$$\Bbb{Z}$_{n}, +_{n}$\rangle$) will not be a generator for $\Bbb{Z}$_{n}. Thus, in our case n = 11 and we know that there are no common divisors of 11 that are less than 11 since 11 is a prime number. That means that all positive integers less than 11 are generators for $\Bbb{Z}$_{11}. Here, we list all generators for $\Bbb{Z}$_{11}: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Each generator will give us the entire set $\Bbb{Z}$_{11} by the definition of a cyclic group, so even though each one forms a subgroup of $\langle$$\Bbb{Z}$_{11}, +_{11}$\rangle$, they all create the improper subgroup so there are no new subgroups being created out of the fact that $\langle$$\Bbb{Z}$_{11}, +_{11}$\rangle$ is cyclic.

**What are the subgroups of $\mathbb{Z}_{11}$?**

A subgroup is defined as a subset H of a group G that is closed under the binary operation of G and that is a group itself. The idea that H is a subgroup of G will be denoted H < G. There are two subgroups that exist for every group, the improper subgroup and the trivial subgroup. The improper subgroup is the subgroup consisting of the entire group G itself. So, the improper subgroup of $\Bbb{Z}$_{11} is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with binary operation +_{11}. The trivial subgroup is the subgroup {e}. In our case, the trivial subgroup is {0} since e = 0 here. There are no other subgroups of $\Bbb{Z}$_{11} because every number that is less than 11 is a generator of $\Bbb{Z}$_{11} and so they all create the improper subgroup.

**What are the cosets of $\mathbb{Z}_{11}$?**

Since we have already shown that the only subgroups of $\mathbb{Z}_{11}$ are the trivial subgroup and the improper subgroup, there only exists one coset, which is the improper subgroup itself.

**Is $\mathbb{Z}_{11}$ cyclic?**

There exists another way of writing out the improper subgroup, H = { $a^n | n$ $\epsilon$ $\Bbb{Z}$}, that is called the cyclic subgroup <a> of G, generated by a. Since H is a subgroup, it means that it is also a group and thus we can define a cyclic group G = <a> where a $\epsilon$ G and G = {$a^n | n$ $\epsilon$ $\Bbb{Z}$} such that a is a generator of G. We now show that $\Bbb{Z}$_{11} is a cyclic group.

*Claim:* $\mathbb{Z}_{11}$ is a cyclic group.

*Proof:* Let a = 1 since 1 is in $\mathbb{Z}_{11}$. So, we want to show that $\Bbb{Z}$_{11} = {$1^n | n$ $\epsilon$ $\Bbb{Z}$}. The group that we produce here is $\lbrace0, 1^1 = 1, 1^2 = 2, 1^3 = 3, 1^4 = 4, 1^5 = 5, 1^6 = 6, 1^7 = 7, 1^8 = 8, 1^9 = 9, 1^{10}=10\rbrace$ which is the whole group $\mathbb{Z}_{11}$. Thus, $\mathbb{Z}_{11}$ is generated by the single element 1 and is a cyclic group. QED

**What are some isomorphisms of $\mathbb{Z}_{11}$?**

In Fraleigh's book, Theorem 6.10 states that if G is a cyclic group with generator a and has finite order n, then G is isomorphic to $\langle\mathbb{Z}_n, +_n\rangle$. Thus, all cyclic groups with finite order 11 are isomorphic to $\mathbb{Z}_{11}$.