Does there or does there not exist $A$ of infinite order in $\{ A \in GL_2(\mathbb{R}) : A^T = A^{-1} \}$?    I know that elements of the form $$A=\begin{bmatrix}\cos(\theta) &amp; \sin(\theta)\\\pm\sin(\theta) &amp; \pm\cos(\theta)\end{bmatrix}$$ are in this group, but do not know how to calculate their order.
121501
$H$ is the set of the matrices $A$ of the form $$A= \begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}$$  where $0\leq \theta &lt; 2\pi$  is a group with respect to matrix multiplication.  Question part (1) Prove by induction that  for any $\theta$ and any positive integer $n$,  $$\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{n} = \begin{pmatrix} \cos(n\theta) &amp; \sin (n\theta) \\ -\sin (n\theta) &amp; \cos (n\theta)\end{pmatrix}$$   When $n =1$ we have $$\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}$$   and clearly this holds for the base case.   Now I'm going to assume true when $n=k$ and try and prove for $k+1$. So we have   $$\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{k}\begin{pmatrix} \cos(\theta) &amp; \sin (\theta) \\ -\sin (\theta) &amp; \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) &amp; \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) &amp; \cos (k\theta + \theta)\end{pmatrix}$$   Now I'm sure I can cheese this, but is there some clever way to pull this apart? I want to just be able to say something like the determinant of the left side is $1^{k+1}$ and its the same on the right ...  Question part (2) Find an element of $H$ of order 5.  Here I'm totally lost, I seem to think that every element is of order 1 cause the determinant is 1.  Question part (3) Find an element of $H$ of infinite order.  Again I'm lost; I want to somehow make the determinant 0 for this but it can never be 0 for any $\theta$.
86847
I've just left an exam of algebraic structures, and one of the questions asked to compute the order of $SL_2(\mathbb{Z}/ 5\mathbb{Z})$ (the 2x2 matrices with determinant $1$ over the field $\mathbb{Z}/ 5\mathbb{Z}$). The answer should be $120$, and a friend of mine could achieve that, but by methods of raw counting. My question is: Is there a way to compute the order of $SL_2(\mathbb{Z}/ n\mathbb{Z})$ for all $n$?  OBS: Although $120=5!$, I believe this is purely a coincidence.
41242
I need help to prove this result:  "Let $G$ be a group such that the intersection of all its subgroups other than $\{1\}$ is a subgroup different from $\{1\}$. Then all its elements have a finite order".  I know I must think of an element $g$ of infinite order. That will imply that every subgroup has infinite order (because it will contain an element like $g^k$). After this, I don't know which step I can take. Can someone help?
8302
My reference for learning abstract algebra says that in an abelian group $G$ the product of two elements (say $a$ with order $n$ and $b$ with order $m$) has order $mn$ if $gcd(n,m) = 1$.   What I don't understand is where the $gcd(m,n)=1$ comes into use in proving this fact? For example, $ a^{mn} = (a^n)^m = (e)^m = e$ (identity element). A similar argument for $b^{mn}$ together with the fact that $G$ is abelian is enough to show that $(ab)^{mn} = e$. From here, how do I conclude for an integer $r &lt; mn$ that $(ab)^r \neq e$  Also, are there any other conclusion that can be drawn if $gcd(n,m) \neq 1$ ?  Thanks for any feedback in advance !
22719
$F$ is a finite field of order $q$. What is the size of $\operatorname{GL}_n(F)$ ?  I am reading Dummit and Foote "Abstract Algebra".  The following formula is given:  $(q^n - 1)(q^n - q)\cdots(q^n - q^{n-1})$.  The case for $n = 1$ is trivial.  I understand that for $n = 2$  the first row of the matrix can be any ordered pair of field elements except for $0,0$. and the second row can be any ordered pair of field elements that is not a multiple of the first row.  So for $n = 2$  there are $(q^n - 1)(q^n - q)$ invertible matrices.  For $n\geq 3$, I cannot seem to understand why  the formula works.  I have looked at Sloane's OEIS A002884.  I have also constructed and stared at a list of all $168$ $3\times 3$ invertible matrices over $GF(2)$.  I would most appreciate a concrete and detailed explanation of how say $(2^3 - 1)(2^3 - 2)(2^3 - 2^2)$ counts these $168$ matrices.
4128
On Dummit's Abstract Algebra on p. 128, it says:     "It is easy to see that $(1 2)(3 4)$ ... does not commute with any non-identity element of odd order in $A_5$."   But I don't find it easy.  Any help, please?
12530
Prove that any finite group $G$ of even order contains an element of order $2.$[ Let $t(G)$ be the set $\{g\in G|g\neq g^{-1}\}$. Show that $t(G)$ has an even number of elements and every nonidentity element of $G-t(G)$ has order $2$.]   My proof:  Observe that for any $g\in t(G)$ there exists $g^{-1}(\neq g)\in t(G)$ since every element in a group has an inverse. Thus if there are $n$ elements that belong to $t(G)$ we must have another $n$ elements that belong to $G$. Hence, $t(G)$ has an even number of elements. Now any $g\in G-t(G)$ must also have inverse $g^{-1}(=g)\in G-t(G)$ since all $g\neq g^{-1}$ belong to $t(G)$, so all the elements except the identity in $G-t(G)$ have $g=g^{-1}$ thus $g^2=id$ i.e., every element of $G-t(G)$ has order $2.$ Therefore, we can conclude that there is atleast one element of order $2.$ in $G$.     Can anyone check my proof? Thank you.
79181
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).  I've already constructed two non-isomorphic total orderings. They are $$ 1 &lt; 2 &lt; 3 &lt;\ldots\\ 1 &gt; 2 &gt; 3 &gt;\ldots $$ where "$&lt;$" and "$&gt;$" are  strict total ordering relations.  Thanks in advance for your help.
44474
Let $G$ is a group and $a, b \in G$. If $a$ has order $6$, then the order of $bab^{-1}$ is...  How to find this answer? Sorry for my bad question, but I need this for my study.
120865
Show that a group with $|G| = 33$ contains an element of order $3$.   Using Lagrange's theorem, the possible orders are $1$,$3$,$11$, and $33$ (from the factors of $33$).  If there is a subgroup containing an element with an order of $33$, then we are done.   There is only one identity element with an order of $1$.   I am currently left with elements of orders $3$ or $11$. What is the best way to show that any element of order $11$ is a contradiction?
77239
Let $a,b$ be two group elements of finite order that commute.   What can be said about the order of $ab$?   I thought that $|ab| = \text{lcm}(|a|,|b|)$. My proof was that $(ab)^n = a^n b^n =e$ if and only if $a^n = b^n = e$ if and only if $|a|,|b|$ both divide $n$. The smallest $n$ such that both orders divide it is the least common multiple of $|a|$ and $|b|$.  By chance I came across . It has an upvote so it's clearly correct. (?)  But it contradicts what I think: It's clear that   $$ (ab)^{\text{lcm}(|a|,|b|)} = e$$  hence $|ab| \mid \text{lcm}(|a|,|b|)$.  It seems to me that this is saying more than $|ab| \mid |a| |b|$. Is it not?     Now my question is: what is the precisest statement that can be made about $|ab|$?  Is there anything more than $|ab| \mid \text{lcm}(|a|,|b|)$ that can be said about $|ab|$?
97627
I have to calculate the order of the ring $\mathbb{Z}[i]/(1+i)$. This is how far I am: If $a+bi\in \mathbb{Z}[i]/(1+i)$ then there are $n,m\in \mathbb{Z}$ such that $a+bi\equiv 0+ni$ or $a+bi\equiv m+0i $. This means that $\mathbb{Z}[i]/(1+i)=\{x\in \mathbb{Z}[i]: Re(x)=0 \text{ or }Im(x)=0\}$. So the order of this ring has to be infinite. I have a feeling I have something wrong, but cant figure what. Is there something wrong in this calculation? Thanks.
80115
I found this question in Arhangel'skii and Tkachenko's book Topological Groups and Related Structures. The first chapter of the book is devoted to algebraic preliminaries.  The question actually reads:     Give an example of an infinite abelian group all proper subgroups of which are finite.   What I have done is: Every element of this group has finite order, else we could find an infinite proper subgroup, namely the group generated by $x²$ if $x$ has infinite order.  I think this can be strengthened: every element should have a prime order. Although I haven't proved this.  Intuitively this group cannot be and infinite product of smaller groups, because you could take the product of the even group factors and find an infinite proper subgroup.  Well, this is it, a highly non-trivial problem. Thanks in advance.
41139
I'm trying to prove that any group $G$ of infinite order has an infinite number of subgroups.  I think that if the group has an element of infinite order, then it's easy because I can take the groups generated by the powers of this element.  What if it doesn't? Every element generates a cyclic subgroup. Every element belongs to at least one cyclic subgroup (that generated by itself). So the group is the union of its cyclic subgroups. If all these are finite, we would have to have an infinite collection of subgroups anyway.  Is that correct?
26707
Let $G$ be the group of all $n\times n$ matrices with entries of the matrices from the field $\mathbb Z_p$ , $p$ prime, such that determinant of every matrix is not $[0]$ , w.r.t. matrix multiplication ; then what is the order of $G$ ?
50208
Let $\mathbb{F}_3$ be the field with three elements. Let $n\geq 1$. How many elements do the following groups have?   $\text{GL}_n(\mathbb{F}_3)$ $\text{SL}_n(\mathbb{F}_3)$   Here GL is the , the group of invertible n×n matrices, and SL is the , the group of n×n matrices with determinant 1.
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.