What are the orders of the elements in $\mathbb{Z}_n$? If $n$ is prime then 0 has order 1 and the rest order $n$. If $a|n$ then $a$ has order $n/a$, and if $a$ is prime then it has order $n$. So are the orders of the elements divisors of n? Say, we take $\mathbb{Z}_{120}$. Are the orders appearing the divisors of 120?
121785
I have this question:     Let $x, n$ be integers with $n \geq 2$ and $n$ not dividing $x$.  Show that the order o($\bar{x}$) of $x \in Z_n$ is   $o(\bar{x})= \frac{n}{HCF(x, n)}$   I've been thinking about it for ages but I still don't get why. A hint would be appreciated.
43847
Let $\sigma = (1234) \in S_6$. List all elements in $N(\sigma) = \{\alpha\in S_6 \mid \sigma\alpha=\alpha\sigma\}$.   I know this relates to cycles and orders, but I am having trouble finding the order of $N(\sigma)$ and then the list of elements.
53156
If $A$ and $B$ are finite subgroups, of orders $m$ and $n$, respectively, of the abelian group $G$, Prove that $AB$ is a subgroup of order $mn$ if $m$ and $n$ are relatively prime. Definition: $AB=\{ab|a\in A, b\in B \}$   My attemptation: $A$ has $m$ elements, $B$ has $n$ elements. There are $mn$ representations of elements in $AB$. If some of them are equal or there exist $a,a'\in A, b, b'\in B$ such that $ab=a'b'$, then $AB$ has elements less than $mn$, if such elements do not exist then $AB$ has order of $mn$. Therefore we want to show that if $m$ and $n$ are relatively prime then such elements do not exist. Please show me how to do it. I think it has something to do with Lagrange's theorem and properties of prime numbers.
116
Though I've seen several cool axiomizations of $\mathbb{R}$, I've never seen any at all for $\mathbb{Z}$.  My initial guess was that $\mathbb{Z}$ would be a ordered ring which is "weakly" well-ordered in the sense that any subset with a lower bound has a minimal element.  However, after seeing  of a discrete ordered ring, I'm less sure. I made that guess under the impression that the fundamental characteristic of $\mathbb{Z}$ is that every nonzero element has exactly one representation of the form $\pm (1+1+\dots+1)$, but that seems to be shared by every DOR.   Presumably, this definition wouldn't exist if every instance of it was isomorphic to $\mathbb{Z}$, so can someone give me an example of another discrete ordered ring? More to the point, what is a sufficient characterization of $\mathbb{Z}$? (and a proof sketch of uniqueness would be nice)  I'm aware that $\mathbb{Z}$ is pretty easily constructible from $\mathbb{N}$, but I want to use this for a seminar and given the audience I am expecting, I would rather not deal with Peano. (And I guess it feels like cheating to say "$\mathbb{N}$ is a well-ordered rig")
57235
Let $p_n$ be the $n$th prime number in its natural order. We want to show that  $$p_{n+1} \leq p_1\dotsc p_n + 1 \tag{1}$$  I know $p_1\dotsc p_n + 1 &gt; 1$, so it has at least one prime divisor, but the text says that $p_{n+1}$ cannot exceed the smallest of these divisors, but I cannot understand why, especially when I imagine $p_{n+1}$ being very large. Although some examples confirm that this statement is true, I cannot make a connection as to why the $n+1$ prime number cannot exceed the smallest prime divisor. Once this has been shown to be true, $(1)$ immediately follows.
21231
I'm struggling with this. I know it is going to use Lagrange's Theorem.  This is what I have so far.  Suppose $|U_n| = k.$  This implies $a^k = 1$ for all $a$ in $U_n$ and $|a|$ divides $k$.  Now, what can be said about the order of an element $a$? I haven't been able to conclude that it's even with what I have, therefore I continued so: $|\langle a\rangle|$ divides $k$ where $\langle a\rangle$ is the cyclic subgroup generated by $a.$  However, I haven't been able to conclude that $|\langle a\rangle|$ is even.   So I tried using this:  an element $a$ is in $U_n$ iff gcd$(a, n) = 1$.  Any hints?
105135
Find the order of $8/9, 14/5,48/28$ in the additive group of $\mathbb{Q}/\mathbb{Z}$    JUNK  Guessing asking for $8/9 * k = 0 $ (being a whole integer) for smallest integer so for this one is 9 . another guess would be the smallest element of $Q/Z$ being something like $9/8$  $14/5$ would be 5  $48/28=12/7$ would be 7?
90939
In $\mathbb{Z}\left [ x \right ]$, let $I = \left \{ f\left ( x \right ) \in \mathbb{Z}\left [ x \right ]: f\left ( 0 \right ) \text{ is an even integer} \right \}$   In fact, $I=\left \langle x,2 \right \rangle$  How do I show that the order of $\mathbb{Z}\left [ x \right ]/I$ is 2?
101026
Let $G$ be a cyclic group of order $m$. What is the order of $\text{Aut}(G)$?  I want to know the proof as well (elementary if possible). I would still accept the proof if one answers with $m = p$, a prime. Or on top of that, I would accept the answer with extra assumption: $q \equiv 1$ mod $p$ with another prime $p$.
62815
Let $a$ and $b$ be elements of a group, with $a^2=e, b^6=e$ and $ab=b^4a.$ Find the order of $ab$ and express the inverse in each of the terms $a^mb^n$ and $b^ma^n.$  Just want to cross check my solution. Would be very grateful for the complete solution.
7918
Consider the quotient group $\mathbb{Q}/\mathbb{Z}$ of the additive group of rational numbers. Then how to find the order of the element $2/3 + \mathbb{Z} $ in $\mathbb{Q}/\mathbb{Z}$.
112845
Let $ G $ be an abelian group of order 6.  Then $ G $ has one element of order 6.  And so, $ G $ is cyclic and isomorphic to $ \mathbb Z _6 $.  In general, It is not true that an abelian group of order $n$ has an element of order $n$.  However in the case of $n=6$, it is true. I'm wondering there is any rule of the case (for example, a rule of $n$ that makes an abelian group cyclic).
60570
Studying groups and subgroups I find this question:     Are there subgroups of order $\mathbf 6 \mathbf 5$ in the additive group $(\Bbb Z$,$+)$?   I would answer no, because a subgroups of $(\Bbb Z,+)$ is the multiple of a Natural number $n $ and it has the form: $n\Bbb Z$={$na|n \in \Bbb N, a \in \Bbb Z$} and they have no finite order.  But I'm not sure of this answer.  Could someone explain if I'm wrong and why?  Thanks.  $\mathbf {edit}:$  My question comes because discussing with a course mate he argued that exist Z/65Z in (Z,+), the group module 65 and it  has order 65 so I'm a bit confused.
48825
I want to show that if $g,h\in G$ are group elements with finite coprime orders $m$ and $n$ and $gh=hg$ then the order of $gh$ is $mn$.    We have that  $$ (gh)^{mn}=g^{mn}h^{mn}=1 $$  using $gh=hg$, so $|gh|\leq mn$. On the other hand, $mn=\operatorname{gcd}(m,n)\operatorname{lcm}(m,n)=\operatorname{lcm}(m,n)$, since $m$ and $n$ are coprime. I am not sure how to proceed, I tried assuming that $|gh|&lt;mn$ and arriving at a contradiction, but it did not take me anywhere.
42453
1) Is $\mathbb{Z}/\langle n\rangle$ cyclic and of order $n$? Why or why not? ($\mathbb{Z}/\langle n\rangle$ is defined to be the factor group of $\mathbb{Z}$ determined by $\langle n\rangle$.)  I guess I'm having trouble understanding factor groups (quotient groups make much more sense) and cyclic groups. My textbook isn't the best at explaining what's going on. I just chose some past homework problems that looked related to my problems.  This is my first abstract algebra course and I'm floundering a lot.
104025
I am a student trying to learn some abstract algebra this summer, and I recently proved (as an exercise) that if $G$ is a group where every element has order 2, then $G$ is abelian. I was wondering could we make a similar conclusion about groups where every element has order 3, namely I am asking if $G$ is a group where all elements have order 3, then $G$ is abelian. I think that this is not true, but I cannot think of a counterexample.   The only groups that I can think of which have all elements order 3 are the groups $(\mathbb{Z}/3\mathbb{Z})^n$, but these are abelian. Any help is appreciated. Thanks!
69532
What is the intuition behind the fact that if every element in a group is of order $2$, we have that the group is abelian? I can prove it, but I do not know the intuition behind it.