While doing my homework i find out that the maximum order of an element in $S_3$ is 3 (the element $(123)$) and the maximum order of an element in $S_4$ is 4 (the element $(1234)$)  Can i generalize that and say that the maximum order of an element in $S_n$ is n?
123501
Consider the symmetric group $S_5$. I would like to find how many elements of $S_5$ are of order 5, and how many are of order 6. I would also like to determine what the maximum order of an element in this group would be.  Here is what I have so far: elements of order 5 are the 5-cycles. Elements of order 6 (since a 6-cycle is impossible for 5 elements) must have at least an even cycle and a cycle of length divisible by 3, and 2 + 3 = 5, so elements of order 5 are a combination of 2-cycles and 3-cycles.  How can I find the total count of such elements of order 5 and order 6, and the maximum order? I'm not entirely sure where to go from here.
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?
20035
Let u and v be elements of a commutative group, and suppose that their   orders are r and s, respectively. Show that if r and s are distinct   primes then the order of uv is rs.   I will use the follwing theorem to proof this.  Theorem: Let x be an element of order m in a finite group G. Then $$x^s=1$$ in G if and only if s i a multiple of m.    My attempt  We can here see that $$(uv)^{rs}=[commutative]=u^{rs}v^{rs}=1^s1^r=1$$ By the theorem, this tells us that this is true only if only rs is a multiple of the order of $uv$. Let the order be $ord(uv)=k$. We know that every positive integer can be written as $$a=bm+l, 0&lt;=l&lt;b$$ and since $r$ and $s$ is primes (positive integers), $rs$ can be written as $$rs=km+l, 0&lt;=l&lt;k$$ This gives ut that $$1=(uv)^{rs}=(uv)^{km+l}=1^muv^l=(uv)^l$$ If l>0 then the equation $(uv)^l=1$ contradicts that the positive integer k is the least positive integer for which $(uv)^k=1$. Therefore l=0 and $rs=km$ . We can here see that $rs$ is a multiple of k (and this proof also the theorem).  $$rs=km =&gt; k | rs =&gt; k|r     ORk|s$$ But we know that $r$ and $s$ are distinct primes, so k must be equal to one of the primes.  $$k=r =&gt; (uv)^k=(uv)^r=1v^r=v^r=1$$ And since we know that $ord(v)=s$ the theorem tells us that $r$ must be a multiple of $s$. But since both of them is primes, we get that $r=s$ but this is a contradiction (r and s are distincts as the exercise tells us).   Now, we try with $k=s$. Same as above, we get that $s=r$ which is a contradiction. Neither k=r nor k=s is right since by the theorem we get that $k=r=s$ and we have that $r$ and $s$ are distincts. This tells us that k cannot be equal to r nor s, since this gives us that $r=s$. And since we know (and as i did proof)that  $$rs=km$$, this tells us, as i wrote before, that $$k|rs$$ Since k is neither equal to $r$ or $s$, this tells us that k must be equal to $rs$. Therefore $$ord(uv)=k=rs$$.  Edit: i know that this can be prooved by l.c.m(x,y) but we have not used it so far. That's why i did not proof it with l.c.m(x,y)
113461
I know that $S_n$ is generated by a number of things, like all transpositions, all transpositions of the form $(1\,a)$, the transpositions $(1\,2),(2\,3),\dots,(n-1\, n)$, and  just the two elements $(123\dots n),(1\,2)$.      Suppose $n$ is prime. If you just have $(123\dots n)$ and some arbitrary transposition $(a\, b)$, how does this also generate $S_n$?    Can you somehow get to $(1\,2)$ or reduce it to some other previous case?
21782
If $g\in G$ is an element of maximal order in a finite abelian group $G$ then exists $H\leq G$ such that $G=\left&lt;g\right&gt;\oplus H$  Attempt: Using fundamental theorem I know that $G=C_{n_1}\times\cdots\times C_{n_k}$ where $n_i|n_{i+1}$. With some work I proved $|\left&lt;g\right&gt;|=n_k$, so $C_{n_k}\cong \left&lt;g\right&gt;$. Then $G\cong H\times \left&lt;g\right&gt;$, but this is not an equality.
92746
I have seen this everywhere but how do I show that : if $a$ is an element of order $n$ in a group $G$ then the order of $a^k$ is $\frac {n}{(n,k)}$.  We know, since $a$ has order $n$, then $a^n = 1$
61762
Can anyone please help me in this question:     $(X,\leq )$ is a well-ordered set. $\forall x\in X$, either $x$ is the largest element of $X$ or there exists $y\in X$ such that   (i) $x&lt;y$ and   (ii) if $x_0\in X$ such that  $x\leq x_0\leq y$, then either $x = x_0$ or $x_0 = y$.   Will this be sufficient to start the proof:  If $x$ is the largest element then there is nothing to prove. Suppose $x$ is not the largest element. $$A=\{ z\in X \mid x&lt;z\}\neq \emptyset$$ How do I proceed?
34209
I'm trying trying to prove that an $n$-cycle in some symmetric group has order $n$, but I don't know where to start. Would appreciate some hints.
103229
Let $S$ be a set of natural numbers,such that each element divides $2015^{200}$ but for no two elements $a$ and $b$, $a|b$. Find the maximum number of elements in $S$ .   $2015^{200}=(5\cdot 13\cdot 31)^{200}$  Hence each element is of the form $5^p13^q31^r$ where $p,q,r$ lie between $0$ and $200$, inclusive.The given condition also says that for no two elements $5^p13^q31^r$ and $5^a13^b31^c$,$p\le a,q\le b,r\le c$.  I have tried to guess the solution by fixing the value of $a$ and then letting $b$ and $c$ run through the range of values that they are allowed to take.But I keep getting confused again and again.  I will appreciate any sort of comment,hint or answer.  EDIT:  As has been noted in the comments below,a trivial upper bound is $201^2$. But is this achievable?Can we show that it's not?Because then we may gain some insight.
105853
I'm looking for an elegant way to show that, among non-negative numbers, $$ \max \{a_1 + b_1, \dots, a_n + b_n\} \leq \max \{a_1, \dots, a_n\} + \max \{b_1, \dots, b_n\} $$  I can show that $\max \{a+b, c+d\} \leq \max \{a,c\} + \max \{b,d\}$ by exhaustively checking all possibilities of orderings among $a,c$ and $b,d$.  But, I feel like there should be a more intuitive/efficient way to show this property for arbitrary sums like the one above.
111385
Let G be the multiplicative group $\mathbb{Z}_n^*$ for $n = 2^k$ and $k \ge 3$.  Can we prove that no element has order bigger than $2^{k-2}$ ?  My solution (not really a solution) :  Since $n=2^k$, I thought that this group is cyclic and the generator is 2. I'm not sure if the group order can't be bigger than $2^{k-2}$?
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!
46211
Let $R$ be a total order on set $S$. Prove that if $S$ has a minimal element, than the minimum element is unique.  I have difficulties with proofs. I know any graph of a total order is a straight line, which clearly has a minimal element. How doI tell when something is mathematically proven?  EDIT: this is how I would prove it  Let $x, y \in S$ such that x and y are minimal. Since S is a total order $x$ and $y$ are comparable and either    case 1: one comes after the other and therefore one isn't the minimal case 2: both are the same and $x=y$  I know this isn't a mathy argument and could use help making it more presentable.
95700
I would like to count subgroups of each order (2, 3, 4, 6, 8, 12) of $S_4$, and, hopefully, convince others that I counted them correctly.  In order to do this by hand in the term exam, I need a clever way to do this because there can be as many subgroups of a group of order 24 as $2^{23}$.  Do you know how to do this?   (I would be most grateful if you could tell me what part of the answer to the old question answers my question before voting to close.)
102223
I was reading Nielsen and Chuang's "Quantum Computation and Quantum Information" and in the appendices was a group theory refresher. In there, I found this question:      Exercise A2.1 Prove that for any element $g$ of a finite group, there always exists a positive integer $r$ such that $g^r=e$. That is, every element of such a group has an order.    My first thought was to look at small groups and try an inductive argument. So, for the symmetric groups of small order e.g. $S_1, S_2, S_3$ the integer $r$ is less than or equal to the order of the group. I know this because the groups are small enough to calculate without using a general proof.   For example, in $S_3$ there is an element that rearranges the identity $\langle ABC \rangle$ element by shifting one character to the left e.g. $s_1 = \langle BCA \rangle$. Multiplying this element by itself produces the terms $s_1^2 = \langle CAB \rangle$; and $s_1^3 = \langle ABC \rangle$ which is the identity element, so this element is equal to the order of the group, which is three.   I have no idea if this relation holds for $S_4$ which means I am stuck well before I get to the general case.   There's a second question I'd like to ask related to the first. Is the order or period of any given element always less than or equal to the order of the group it belongs to?
17840
Why $S_4$ is not isomorphic to $D_{12}$? where $S_4$ is symmetric group and $D_{12}$ is dihedral group of order $12$.here number of element are same $S_4$ has $4!=24$ elements and $D_{12}$ has also $2n$ means $12(2)=24$ elements. but i can't find that property which holds in one of this group but another doesn't has that property.please help me thanks in advance.
78456
I have a lot of difficulty understanding this proof we went over in class about classification of groups of order 12.  Let $G$ be a finite group of order $n=p^rm$ where $m \nshortmid p$.  Denote $Syl_p(G)$ as the set of all sylow p-subgroups, and $n_p(G)$ as the cardinality of $Syl_p(G)$.  (1) G has subgroups of order $p^r$. (How do we know this??)  (2) We know all sylow p-groups are conjugate and their number $n_p(G) | m$, by the 2nd and 3rd Sylow Theorems.  (3) We have that $n_p(G) \equiv 1(p). $  Then $n=12=2^2*3^1$ so $n_2(G)|3$ gives $n_2(G) \in \{1,3\}$ and $n_3(G)|4$ so $n_3(G) \in \{1,4\}$ since $2 \not\equiv 1(3)$.  Case 1: $n_3(G) = 4.$ Then the action of G by conjugation on $Syl_3(G)$ gives a homomorphism $f: G \rightarrow S_{Syl_3(G)} \simeq S_4$. $Ker(f)$ consists of $g \in G$ that normalizes all 3-Sylow subgroups. (Why?). Let $P_3$ be a 3-Sylow subgroup. Then the order of $P_3$ is 3. (I don't get that either), and $[G:N_G(P_3)] = 4.$ Thus $P_3 \subset N_G(P_3)$ gives $P_3 = N_G(P_3)$. (I'm lost here...) So $Ker(f) = \cap Syl_3(G) = \{e\}$ (the intersection of the sylow 3-subgroups is trivial. (How did we arrive here...) Thus we conclude that $G$ is isomorphic to a subgroup of $S_4$ of order 12.  To show $G$ is isomorphic to $A_4$, we have that $G$ has 8 elements of order 3: 4 3-sylow subgroups each has 3-1=2 elements of order 3. And $S_4$ has 8 3-cycles meaning f(G) contains all 3-cycles and hence the group they generate, which is $A_4$, so $A_4 \subset f(G)$. But since $|A_4| = 12 = |f(G)$ then $f(G) = A_4$ and $G \simeq A_4$. (Not understanding the first half of the proof, I do not get this part either).  Case 2: $n_3(G) = 1.$ The scope of the question is too large, I will need to post separately to understand the second half, unless the organizational rules of math.stackexchange would require that I post here. In that case, I'll edit the question.  I basically cannot follow the proof because it seems to skip too many steps; it would be helpful if anyone could elaborate case 1 with more reasoning so that I can follow it. Any help would be appreciated; I've spent hours trying to look up and decipher the proof but without any success.