I'm working on a homework question which states:     Show that the real Heisenberg group        H=  \begin{bmatrix} 1 &amp; a &amp; c\\ 0 &amp; 1 &amp; b\\ 0 &amp; 0 &amp; 1 \end{bmatrix}    for a,b,c $\in \mathbb{R}$ is nilpotent.    I've shown the center of $H$ to be matrices of the form \begin{bmatrix} 1 &amp; 0 &amp; c\\ 0 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 1 \end{bmatrix}  and I am now trying to calculate the center, $Z_2(H/Z(H))$, for the ascending central series. I think the center is supposed to be all of $H/Z(H)$, so I am trying to show these cosets are abelian, and I think that is where my issue is. For two cosets, $hZ(H)$ and $gZ(H)$, $(g,h \in H)$ I am confused what this looks like and how to show they commute.   Any help on this is appreciated. I know there are other ways to do this, but I would like to use the ascending central series since that is the way I am learning it. Thanks!  In response to the duplicate, I understand this is the same question, but I think my approach is different. Looking at the other question, they are using a method I am unfamiliar with. I think I can show that the cosets are abelian, but am unsure how to proceed from there. Thanks for the feedback!
123948
This time I want to show that $H(R)$ ( the heisenberg group attached to a conmutative ring $R$ is nilpotent). But I dont know how to proceed, I was thinking in first serch the normal subgroups if the Heisenberg group and then try to figure out the quotient group of that, from there get the centre, etc, but I dont think it would be a good idea (In fact I thought to use Gap, but I dont know how to ask this questions in Sage), Is there a better way to do this? Thanks a lot.   For me   $H(R) = \left \{ \begin{pmatrix} 1 &amp; a &amp; c \\ 0 &amp; 1 &amp; b \\ 0 &amp; 0 &amp; 1 \end{pmatrix}: \ \ a,b,c \in R \right \}$  For a conmutative ring $R$
66646
Let $A_R$ be the finitely generated abelian group, determined by the relation-matrix       $$R := \begin{bmatrix} -6 &amp;  111  &amp; -36 &amp; 6\\ 5  &amp;  -672 &amp; 210 &amp; 74\\ 0 &amp; -255 &amp; 81 &amp; 24\\ -7  &amp;   255       &amp;-81 &amp; -10 \end{bmatrix}$$       Reduce this matrix using Smith Normal Form and determine the isomorphism type of $A_R$.    I know that the Smith Normal Form of this matrix is:  $$\begin{bmatrix} 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 3 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 21 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0  \end{bmatrix} $$  However, this was computed using Maple and I need to understand the method of computing this manually which I am struggling to grasp. Can anyone help?
90167
If $a, b \in G$ commute   , show that  $o(ab) \mid\operatorname{lcm}(o(a), o(b))  $.  (Hint: Let $L=\operatorname{lcm}(o(a), o(b))$ and prove that $(ab)^L =e$.    ??  So I know that  *Suppose a has order n ,and b has order m ; that is, a^n=e and b^m=e   Then, since G is abelian, we have $(ab)nm=^{nm}=(a^nma^{nm})(b^nmb^{nm})=((a^n)^m))(^m\cdot (b^m)^n))=e^n=e$.  Now, the order of ab may be smaller, but it must divide $     nm=\operatorname{ord}(a)\cdot\operatorname{ord}(b)$.*  how iHow I can answer this question?
116675
How to prove that center of $GL_n(\mathbb R)$ is scalar matrices?  Again what can you say about the center of set of all upper triangular matrices and center of set of all diagonal matrices.
1128
Let $G$ be a group. If $H\leq Z(G)$ and $G/H$ is nilpotent, then $G$ is nilpotent. To prove this proposition, first, I have tried to show that $G/Z (G)$ has non-trivial center. I can see that all cosets of $H$ in $Z (G)$ belongs to $Z (G/H)$ but this does not satisfy $G/Z (G)$ has non-trivial center. Can any one give me a hint for the proposition. Thanks.
106253
I have just started learning about Lie algebra in the context of quantum mechanics and got confused with this: Some sources say the generators are $J_0,J_1$ and $J_2$ and some use $J_0,J_+$ and $J_−$. Which set is correct? Or if both are correct what key concept am I missing here? My understanding is that if have certain commutation relations then we know that Lie Algebra is such and such. But if we have two such choices then this understanding falls apart? How do I figure out what is $\mathfrak{su}(2)$ Lie Algebra in general?
20523
The center of a group $G$ is defined as $Z(G):=\{ z\in G : gz = zg, \; \forall g \in G\}$.  The goal is to show that if $\vert G\vert = pq$, where $p$ and $q$ are not necessarily distinct primes then either $G$ is abelian or $Z(G) = \{ e\}$.  I want to suppose that $Z(G) \neq \{ e\}$ and then use the fact that $G/Z(G)$ is cyclic to imply that $G$ is abelian, which is something I have already proven. But how do I show that $G/Z(G)$ is cyclic when I am not certain what exactly $Z(G)$ looks like. I only know that it has at least one non-identity element in it, which will be of order $p$ WLOG, (the case where it is of order $pq$ is trivial).  Any help is appreciated. Thank you.
103191
Does anyone know, where I can find a reference (preferably a book) which says that the general linear group $\text{GL}_{2}(\mathbb{Z})$ is generated by the set  $$\left\{\begin{bmatrix} 1&amp;0\\0&amp;-1\end{bmatrix}, \begin{bmatrix} 0&amp;1\\1&amp;0\end{bmatrix}, \begin{bmatrix} 1&amp;1\\0&amp;1\end{bmatrix}\right\}$$  Thank you in advance.
48351
Let $G$ be a matrix Lie group and let $H$ be a Lie subgroup of $G$. My understanding is that all $g \in G$ are matrices, since $G$ is a subgroup of the general linear group. Then aren’t all elements $h \in H$ matrices, since $h \in G$?  From what I can gather $H$ is a matrix Lie group if $H$ is a closed subset of $G$. But if $H$ is not a closed subset of $G$, why is it not a matrix Lie group? Is the reasoning in the first paragraph wrong?
100999
If $H \leq Z(G) \leq G$, where G is a finite group, $Z(G)$ is its center, and $(G:H) = p$ for some prime $p$, then G is abelian.  Well because H &lt; Z(G) we have that H commutes with all of G. Also |G/H|=p. I was thinking of connecting this result to the fact that if $G/Z(G)$ is cyclic, then $G$ is abelian, but was unable to. Then I thought about using the result that $(G/H)/(Z(G)/H) = G/Z(G),$ but that leads to nowhere. Any suggestions?
1110
I'm currently working my way through Harvard's online abstract algebra lectures (if you're interested, you can find them ).  The lectures come complete with notes and homework problems.  Of course, since I don't actually go to Harvard, I can't hand in the homework assignments to find out if I'm doing them correctly.  So I've decided to try to crowd source the grading of my solutions.  I'll post individual questions as I finish them and wait for comments.  I would like to get critiques of not just my reasoning but the style of my write ups as well.  Also, any alternative approaches to the problem would be welcome.  I've looked for forums dedicated to these kinds of OCW courses but have not been able to find any.  This surprises me.  It seems like, with the advent of these free online educational resources, an online meeting place for those who take advantage of them would be a natural offshoot.  So this might be a bit of an experiment.  Unless I'm the 873rd person to post something like this here.  If that's the case, sorry for being so long winded.  Anyway, on with the question.  This one is assigned in the 4th lecture.     Let $V$ denote the Klein 4-group.  Show that $\operatorname{Aut}(V)$ is isomorphic to $S_3$.   (I've moved my solution below to keep this question from showing up in the unanswered list.)  Thanks...
24224
I want to prove, that $GL(n,K)$ is open in $M(n,K)$, where ($K=\{\mathbb{R},\mathbb{C},\mathbb{H}\}$.  For the prove I don't want to use that the determinant is continous. Alternativly I assume a submultiplicative norm.  First step: Define $U_A:=\{B\in M(n,K):\Vert A-B\Vert&lt;\frac{1}{\Vert A^{-1}\Vert}\}$ and show that it is a neighbourhod of $A$ in $M(n,K)$.  Second step: I show that the series $\sum_{k\ge0}(1-A^{-1}B)^kA^{-1}$ converges absolutely and the limit is inverse to $B$.  For the second step I should use the Neumann series, right? But I have some problems with the first step. Any help?
77052
I'm solving this question in Hungerford's book:     8​. Let $G$ be the multiplicative group generated by the real matrices $a=\begin{pmatrix}2&amp;0\\0&amp;1\end{pmatrix}$ and $b=\begin{pmatrix}1&amp;1\\0&amp;1\end{pmatrix}$. If $H$ is the set of all matrices in $G$ whose (main) diagonal entries are $1$, then $H$ is a subgroup that is not finitely generated.   I've already proved that $H$ is a subgroup of $G$ and every element of $H$ is of the form:   $$         \begin{pmatrix}          1 &amp; x \\         0 &amp; 1 \\         \end{pmatrix} $$  I thought this question really weird, $H$ is a subgroup of a group generated by $a$ and $b$ but $H$ itself can't be generated by $a$ or $b$, how can be possible?  My question is how to show $H$ is infinitely generated? I didn't find any good candidates to generate $H$.  Thanks in advance
51151
I computed the homology groups of the torus, can someone tell me if this is correct? The calculation, not the result that is. Thanks!  The cells of $T^2$ are $e^0, e^1_a, e^1_b, e^2$  The chain groups are  $$ C_0(T^2) = \{ k e^0 |&#160;k \in \mathbb{Z} \} = \mathbb{Z}$$  $$ C_1(T^2) = \{&#160;k_1 e^1_a + k_2 e^1_b |&#160;k_1 , k_2 \in \mathbb{Z} \} = \mathbb{Z} \oplus \mathbb{Z}$$  $$ C_2(T^2) = \mathbb{Z}$$  $$ C_k(T^2) = \{ 0 \} , k &gt; 2$$  Now the homology groups:  $$ H_0(T^2) = \ker \partial_0 / im \partial_1 = \mathbb{Z} / 0 = \mathbb{Z}$$  where $im \partial_1 = 0$ because there is no chain in $C_1(T^2)$ whose boundary is a zero-chain in $C_0(T^2)$. (Is this reasoning correct?)  $$ H_1(T^2) = \ker \partial_1 / im \partial_2 = \mathbb{Z} \oplus \mathbb{Z}$$  where $\ker \partial_1 =  \mathbb{Z} \oplus \mathbb{Z} $, i.e. again everything maps to zero because there is no element in $C_1$ whose boundary maps to an element in $C_0$.  $im \partial_2 = 0$ again because there is no element in $C_2$ whose boundary is an element of $C_1$.  I don't want to use Hurewicz to do $H_1$.  $$ H_2(T^2) = \mathbb{Z}$$ using similar arguments as above.  Thanks for your help.  Edit  I posted the computations as an answer below. I got two up votes but I don't know by who so I'm not yet sure I can trust them...
92539
I would like to know whether the $GL^{+}_n(\mathbb{R})$ the set of all invertible matrices with positive determinant is simply connected or not? I guess it is not simply connected but that is just a guess only, I do not know how to prove that, i.e how to show that its fundamental group is non-trivial.  Well, I can rigorously prove that this is connected and hense path connected as it is Lie Group. Could any one rigorously tell me how to approach this kind problem and solve them from the basic knowledge of fundamental group or some other way? So basically I need some result or tools by which I can compute fundamental groups of all known classical matrix Lie groups.
72287
Let $G$ be a group and let $G'$ denotes its commutator subgroup, that is the group generated by all elements of the form $g^{-1}h^{-1}gh$.  Is the following true:      If $G/G'$ is cyclic, then G is abelian.   Recall that the claim is true for the $G$ modulo the center of $G$.  See
30139
Assume that $N$ is a normal subgroup of $G$ such that $N \cap G'=\{e\}$.  Prove that $N \subset Z(G)$.   Note 1: $Z(G)$ is the center of the group $G$, and $G'$ is the set of all commutators of $G$.  Note 2: Assume that $n \in N$. We want to show that for all $g \in G,$  $ng=gn$. But, I don't know how to use the fact that $N \cap G'=\{e\}$ to reach this.    Any hints? suggestions? answers?