I have an exercise which involves groups. It's probably common, but I don't know how to solve it. It sounds like this:      Let (G, $\cdot$) be a group and x, y $\in$ G. Show that ord(xy) = ord(yx).   where ord(x) is the order of x. I just tried to write $xy\cdot xy \cdot \space ...\space\cdot xy = x \cdot (yx) \cdot (yx) \cdot\space ...\space \cdot (yx) \cdot y$, letting x and y outwards. I don't know if this is a good step and I don't know how to continue. I would be glad if someone could help me.
7005
How do I prove that if $a$, $b$ are elements of group, then $o(ab) = o(ba)$?   For some reason I end up doing the proof for abelian(ness?), i.e., I assume that the order of $ab$ is $2$ and do the steps that lead me to conclude that $ab=ba$, so the orders must be the same. Is that the right way to do it?
118133
Suppose $G$ is a group generated by elements $x$ and $y$ where $xy^2 = y^3x$ and $yx^3 = x^2y$ What can you prove about $G$?  I've just been playing around with the relations but I can't seem to get anywhere. It seems as if I'm going in circles. I don't know what to look for in terms of how the equations should look in order to give me an insight into the structure of the group.  So given a group presentation, are there any general strategies of what to look for in the relations (as in how to rearrange them) in order to obtain insight about the group?
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 $p$.
54985
I've figured out that if I know $G$ is not cyclic, then it for any $a \in G, o(a) \neq 4$ (or the order of any element in group $G$ is not 4).  I know ahead of time that the elements in the group ($\forall x \in G$) must have order $o(x)=k$ where $0 &lt; k \leq 4$ where $k \in \mathbb{Z}$, so $k = 1, 2, 3, 4$.  If $G$ not cyclic, then we know $k \neq 4$ so we have $k = 1, 2, 3$ left.  I know how to show that $k \neq 3$ but I am not sure if I am supposed to do it for all possible $k$ because obviously if our order was higher (let's say $n$) then it would get messy. It ends up being in this case that $k = 2$ is good and $k = 1$ is trivial. I have looked up online and it says using Lagrange's Thm we know it has to be $k = 1, 2$ since $3 \nmid 4$, but I cannot use that theorem as we have not have learned it in class. How can I show that $k = 1$ or $2$ in another way?  However working from using the result that if we have $x \in G$ and $H = \langle a \rangle$ and knowing that $\left\vert{H}\right\vert=1$ or $2$, then $x^2=1$ for either case. Then I am done right?  What is another, cleaner, better way of trying to answer the question in the title?  I apologize if my formatting is poor since I am new at LaTeX and extraordinarily bad at algebra it seems. Thank you in advance.
38649
Let $G$ be a group and $g,h \in G$. I need to prove that if $g$ and $h$ commute and their orders are coprime, then $|gh| = |g||h|$, that is, the order of their product is the multiple of their orders.  Since $gh = hg$, then $(gh)^{ lcm(|g|, |h|)} = e$, so, $|gh|$ divides $lcm(|g|, |h|)$. Since $\gcd(|g|, |h|) = 1$, then $lcm(|g|, |h|) = |g||h|$.  So, $|gh|$ divides $|g||h|$.
35250
How can I show that every group of order $4$ is abelian?  Let's denote $e,a,b,c$ as the four elements of the group.  Since $e$ is identity, we have $e*x=x*e$ for every $x$:  $$\begin{array}{|c|cccc|} \hline     &amp; e &amp; a &amp; b &amp; c \\\hline   e &amp; e &amp; a &amp; b &amp; c \\   a &amp; a &amp;   &amp;   &amp;   \\   b &amp; b &amp;   &amp;   &amp;   \\   c &amp; c &amp;   &amp;   &amp;   \\\hline \end{array}$$  Now $a*a=$? We have several possibilities.  If I choose $a*a=b$ I can show $a*b=a*(a*a)=(a*a)*a=b*a$.   But so far I have shown only that $a$ and $b$ commute, there are also other pairs. And I have only discussed the choice $a*a=b$, there are also other possibilities. Is there a simpler way to do this?
15672
Suppose $G=\{a_1,...,a_n\}$ is a finite abelian group, and let $x=a_1a_2\dotsm a_n$.   Prove that if there is more than one element of order $2$ then $x=e$.   What I've done so far: (#1 is just for illustration, There should be at least $3$ such elements)   $G$ has an even number of elements with order $&gt;2$ (They are paired so: $(a,a^{-1})$ ). And then there are $e$ and $\{b|o(b)=2\}$. Also, if $G$ has at least one element of order 2 then $|G|$ is even (Lagrange). Thus $|\{b|o(b)=2\}|$ must be odd. If there is exactly one element $b$ of order $2$ then $x=b$. That's because $G$ is abelian, and we can write: $x=e\cdot b\cdot(a_1 a_{1}^{-1}\dotsm a_n a_{n}^{-1})=b\cdot(e\dotsm e)=b$. If there are exactly $3$ such elements $a,b,c\in G$, then, as shown above, $x=abc$ and thus $x=e$ (That's because $(abc)^2=a^2b^2c^2=e$ which means $abc\in \{a,b,c,e\}$. If $abc=a$ then $b=c^{-1}=c$ which is false, same goes for $abc=b,abc=c$.)   I haven't managed to find a reason why the claim must hold for $5, 7, ...$ (i.e. All other odd integers). Does anyone have an idea?  Thanks in advance. Please excuse my English.  edit: Not a canonical answer but a simple one (if such even exists).
96801
Let $G$ and $K$ be groups and $f:G\rightarrow K$ be a group   homomorphism. Suppose $g\in G$ has finite order $n$. Prove that   $\mathrm{ord}(f(g))$ is finite and divides $n$.   How can I fix this proof?  We know $g^n=e$. Since $f$ is a homomorphism, then $[f(g)]^n = f(g^n)=f(e)=e$. Let $k\mid n$ and $j\mid n$. Then we can choose the smallest $k$ such that $[f(g)^k]^j=[f(g)]^n$. Thus, $\mathrm{ord}(f(g))=k$ where $k\mid n$.  This logic seems shaky. How can I improve it? Thanks!
40122
Let $G=\langle a\rangle$ be a cyclic group of order $n$. Show that, for every divisor $d$ of $n$, there exists a subgroup of $G$ whose order is $d$.  This time I have no approach. I haven't found any relation between the divisors of $n$ and the order of the subgroups of $G$. How would approach this?
58037
Lets view groups as consisting of a binary operation, a distinguished element $e$, and unary operation $x \mapsto x^{-1}$. Then the group axioms can be stated as follows.   $(xy)z=x(yz).$ $xe=ex=x.$ $xx^{-1}=x^{-1}x=e.$   Supposing we dropped axiom 2 and forgot about $e$, we would need to rephrase 3. One possibility would be  3'. $xx^{-1}y=x^{-1}xy=yxx^{-1}=yx^{-1}x=y.$  So my first question is, can 3' be shortened and simplified?  Now suppose we dropped associativity, as well. Then 3' needs to be further rephrased. One possibility would be  3''. $(xx^{-1})y=(x^{-1}x)y=y(xx^{-1})=y(x^{-1}x)=y,$ and $x(x^{-1}y)=x^{-1}(xy)=(yx)x^{-1}=(yx^{-1})x=y.$  Again, my question is, can 3'' be shortened and simplified?
15830
Let $(G,\cdot)$ be a group show that  A) $G$ is abelian  B) For all $x,y\in G: (xy)^{-1}=x^{-1}y^{-1}$  C) For all $x,y\in G: (xy)^{2}=x^{2}y^{2}$  D) There existst an $n\in \mathbb{Z}$ such that for all $x,y\in G$  $\displaystyle (xy)^{n-1}=x^{n-1}y^{n-1}$ and $\displaystyle (xy)^{n}=x^{n}y^{n}$ and $\displaystyle (xy)^{n+1}=x^{n+1}y^{n+1}$  are equivalent.    I showed $A\iff B,A\iff C$ and $A\iff D$. That was not really hard, except maybe $D\Rightarrow A$. I would like to do a circular argument, but i have no idea how.  Any hints?  Thanks in advance!
10580
Let $G$ be a finite, nonempty set with an operation $*$ such that   $G$ is closed under $*$ and $*$ is associative Given $a,b,c \in G$ with $a*b=a*c$, then $b=c$. Given $a,b,c \in G$ with $b*a=c*a$, then $b=c$.   I want to prove that $G$ is a group, but I don't know how to show that there exists an identity $e\in  G$ such that $e*x=x$ and $x*e=x$ $\forall x \in G$. I also don't know how to show that $\forall$ x $\in G$ there exists a $y \in G$ such that $y*x=e$ and $x*y=e$. How do I do this?
108895
Let $G$ be a group and $A$ &amp; $B$ be two sets s.t. $G$ acts transitively on each of $A$ &amp; $B$. Choose some $\alpha$ and $\beta$ in $A$ &amp; $B$ respectively then prove that if $G=G_\alpha G_\beta$ then $G=G_x G_y$ for all $x$ $\in A$ &amp; $y$ $\in B$   where $G_x$ is stabilizer of $x$ in $G$ and similarly $G_\alpha$ be the stabilizer of $\alpha$ and $G_y$ is the stabilizer of $y$.  I know we have to choose some elements and apply some manipulative trick. it is hardly 2 liner but just can't think of right elements. help!
105562
Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$  Question: Deduce that if $G$ is a finite $p$-group, for some prime $p$, the number of subgroups of $G$ that are not normal is divisible by $p$.  Comments:    The subgroups of $G$ that are normal have the property that $g\cdot H=gHg^{-1}=H$ The question is equivalent to showing $p$ divides $\left|G\right|-\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$ $p$-group: $\forall g\in G,|g|=p^k$ for $k\in\mathbb{N}^+$ I don't know where to start with this problem, been thinking about it for a while to no avail, I feel that there are too many definitions for me to consider when finding a solution. My problems I have considered and not been able to answer are: what is the order of $|G|$? I think it must be of order that is divisible by $p$, hence the question becomes show $p$ divides the order of $\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$. How do you compute the order of this set? It seems like quite a standard problem, so I would really appreciate it if I could be directed to more information about whatever problem it is.
107978
I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side. Please guide me how to do it further.     $1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.   Sol:  $P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.  $P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.  $P(1): 3$.  Hence it is true for $n=n_0 = 1$.  Let it be true for $n=k$.  $P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.  We have to prove  $P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.  Taking LHS: $$\begin{align*} 1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &amp;= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\ &amp;= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\ &amp;= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\ &amp;= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\ &amp;= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right]. \end{align*}$$
115187
Let $G$ be a group such that for all $x\in G$ we have $x^2=e$. Show that if $G$ is finite then the order of $G$ is $2^n$.  Here is the solution I have seen in a book.      If G is finite, it can be considered as a vector space over the field $\Bbb{Z}/2\Bbb{Z}$, and then necessarily is finite dimensional, which gives $G$ as a vector space isomorphic to $(\Bbb{Z}/2\Bbb{Z})^n$.   I don't understand why $G$ can be considered as a vector space over the field $\Bbb{Z}/2\Bbb{Z}$. Can, someone, explain this? Please.
68319
My professor assigned this for homework but I don't understand how to connect the dots. He suggested using the fact that $\gcd (x,y) \cdot \operatorname{lcm} (x,y) = xy$ but I'm not sure how that's relevant.