If $G = \begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$)   and assume G is a group under matrix multpication  Assume that K = ($\begin{pmatrix} s &amp; 0 \\ 0 &amp; 1 \end{pmatrix}: s\in\mathbb{R}, s \neq 0$) is a subgroup of G. Prove that K is not a normal subgroup of G.  I want to prove this by showing the left cosets and the right cosets are not the same.  So if we take $g=\begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix}$  then $gK=\begin{pmatrix} as&amp; b \\ 0 &amp; 1 \end{pmatrix}$  And if we take, $g'=\begin{pmatrix} a' &amp; b' \\ 0 &amp; 1 \end{pmatrix}$ then $Kg'=\begin{pmatrix} sa' &amp; sb' \\ 0 &amp; 1 \end{pmatrix}$  How do i show that $gK \neq Kg'$, Is it sufficient to say that $Kg'$ has two columns that depend on s where as $gk$ only has one column that depend on S. So they are not equal
121461
If $G = \begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix} a,b\in (\mathbb{R}) : a \neq 0$)   and assume G is a group under matrix multpication  Assume that K = ($\begin{pmatrix} s &amp; 0 \\ 0 &amp; 1 \end{pmatrix}: s\in\mathbb{R}, s \neq 0$) is a subgroup of G. Prove that K is not a normal subgroup of G.  I have done some working out on this question which is the following:  $g^{-1}=\begin{pmatrix} a^{-1}&amp;-a^{-1}b\\0&amp;1\end{pmatrix}$   and so  $$gkg^{-1}=\begin{pmatrix} a&amp;b\\0&amp;1\end{pmatrix}\begin{pmatrix}sa^{-1}&amp;-a^{-1}bs\\0&amp;1\end{pmatrix}$$  then we get $$gkg^{-1}=\begin{pmatrix}s&amp;-bs+b\\0&amp;1\end{pmatrix}$$  What do I do from here, Why is $$gkg^{-1} \notin K$$ Can't I make b=0 and then it would be in K? or am I missing some key idea?
38784
Let $G$ be a finite group of order $mn$ with $\gcd(m,n) = 1$ and $H$ be a normal subgroup of order $m$. Prove that it is the only subgroup of order $m$. This is somewhat different from this question:  and I don't think we are allowed to use homomorphisms(it is in the next section) I let $K$ be a subgroup of order $m$. If $K = H$, I am done. Otherwise, I supposed $K \neq H$. Based on previous results, I know that $HK = \{hk | h\in H, k \in H\}$ would be another subgroup of $G$ if $H$ is normal. We would have that $\exists k \in K$ such that $k \notin H$ or $\exists h \in H$ such that $h \notin K$. In either case, $|HK| &gt; m$ which may cause problems with some theorems of number theory and Lagrange's theorem but I am unsure of the specifics. I don't know if I can say more about $|HK|$ or whether I am even on the right track. Thoughts?
33980
Given a finite group $G$ and its subgroups $H,K$ such that $$G \times H \cong G \times K$$ does it imply that $H=K$.  Clearly, one can see that this doesn't work out for all subgroups. Is there any condition by which this can remain true.
115544
Could someone please explain something about the following proof? It is based on the proof ().     If $H\leq G$ and $[G:H]=2$, prove that $H$ is normal in $G$.   Proof:  If $[G:H]=2$, then there are two distinct left cosets $H$ and $G\setminus H$. This is because left (right) cosets partition $G$. Let $x\in G$.  Case 1: $x\in H$. Then $xH=H=Hx$.     I don't understand where $H=Hx$ comes from. Is it similar to the proof for "if $x\in H$, then $xH=H$"?   For Case 2: $x\notin H$, the proof for $xH=G\setminus H$ is similar to that of "if $x\in H$, then $xH=H$" as well?   The proof I am basing this on is using "if right and left cosets are equal, then $H$ is normal", but can the same proof be applied for when index is only known and equal to $2$?
98497
So I tried looking everywhere but I'm really lost with this proof. I need to to show that:  Let $H$, $K$ be normal subgroups of a group $G$, satisfying $H\cap K = \{e\}$. Let $h\in H$, $k \in K$. (a) Prove that the commutator $w = hkh^{-1}k^{−1}$ satisfies $w \in H$ and $w \in K$. Then conclude that $w = e$ (b) Prove that $hk = kh$, for all $h \in H$, $k \in K$.
10989
Question: Let $G$ be a group. Prove that $Z(G)$ is a subgroup of $G$.   If I want to show that $Z(G)$ is a subgroup of $G$ that means I have to show that it is closed under group operation?  Here is my attempt.   Let $a,b$ be elements in $Z(G)$ and $x$ be an element in $G$. Then $ax=xa$ which is under group multiplication commutative and under inverse $(a^{-1}) x=x(a^{-1})$. And hence $(a^{-1})bx= (a^{-1})xb=x (a^{-1})b$ which is under group operation so $(a^{-1}),b$ are elements in $Z(G)$ thus a subgroup of $G$.  Really appreciate if anyone can help me by directing me if my attempt is not good. Thanks in advance.
46025
Let $A,B &lt; G$ be two subgroups of some group $G$, then I have a question on the proof of the following: $$  |AB| = \frac{|A||B|}{|A \cap B|}. $$ Proof: Let $D = A \cap B$, arrange $A$ and $B$ in left cosets and right cosets regarding $D$ (which is also a subgroup) $$  A = D \cup s_2 D \cup s_3 D \cup \ldots \cup s_n D, \quad   B = D \cup D t_2 \cup D t_3 \cup \ldots \cup D t_m $$ Now the fact that $\frac{|A||B|}{|D|}$ are distinct results from the following equations: If $$  s_{\alpha}D D t_{\beta} = s_{\alpha_1} D D t_{\beta_1} $$ then $$  s_{\alpha_1}^{-1} s_{\alpha} D = D t_{\beta} t_{\beta}^{-1} $$ As the first member of the last equation represents an element of $A$ while the second member represents an element of $B$, it results that the last equation implies $\alpha_1 = \alpha$ and $\beta_1 = \beta$.  The last step is not clear to me, why does $\alpha_1 = \alpha$ and $\beta_1 = \beta$ follows? Is it the case that if $aD = Db$ for $a \in A$ and $b \in B$ that $a=b=1$ follows? I don't believe this and cannot prove it either? Could someone please explain the last step?
14418
I'm reading through G&amp;P's Differential Topology book, but I hit a wall at the end of section 4. There is a result stating     The set $X=\{A\in M_{m\times n}(\mathbb{R}):\mathrm{rk}(A)=r\}$ is a submanifold of $\mathbb{R}^{m\times n}$ with codimension $(m-r)(n-r)$.    There is a suggestion: Let $A\in M_{m\times n}(\mathbb{R})$ have form $$ A=\begin{pmatrix} B &amp; C \\ D &amp; E\end{pmatrix} $$ where $B$ is an invertible $r\times r$ matrix. Then right multiply by $$ \begin{pmatrix} I &amp; -BC^{-1} \\ 0 &amp; I \end{pmatrix} $$ and show $\mathrm{rk}(A)=r$ iff $E-DB^{-1}C=0$.  I multiplied out and got the matrix $$ M:=\begin{pmatrix} B &amp; 0 \\ D &amp; E-DB^{-1}C\end{pmatrix}. $$  Since I multiplied by a nonsingular matrix, I know that $\mathrm{rk}(A)=\mathrm{rk}(M)$. If $E-DB^{-1}C=0$, then  $$ M=\begin{pmatrix} B &amp; 0 \\ D &amp; 0 \end{pmatrix} $$ has rank $r$, so $A$ has rank $r$. For the converse, if $A$ has rank $r$, then $M$ has rank $r$, so by performing row operations, $M$ is row equivalent to a matrix of the form $$ \begin{pmatrix} I_r &amp; 0 \\ 0 &amp; 0 \end{pmatrix}. $$ This would imply $E-DB^{-1}C$ is row equivalent to $0$, and I think this implies $E-DB^{-1}C=0$.  My main concern is then, how does this approach imply $\mathrm{codim}(X)=(m-r)(n-r)$? Is there some special map I can apply the Preimage Theorem to?  Thank you.
8067
"Let $G$ be a group, and suppose $G=H \cup K$, where $H$ and $K$ are subgroups.  Show that either $H=G$ or $K=G$."  Let $h \in H$ and $k \in K$.  Then $hk \in H$ or $hk \in K$ (since every element of $G$ is in either $H$ or $K$).  If $hk=h'$ for some $h' \in H$, then $k=h^{-1}h'$, so $k \in H$.  If $hk=k'$ for some $k' \in K$, then $h=k'k^{-1}$ so that $h \in K$.  If for all $h \in H$ we have $h \in K$, or if for all $k \in K$ we have $k \in H$, then $H \subseteq K$ or $K \subseteq H$.  Then since $G=H \cup K$, we must have either $H=G$ or $K=G$.  I'm not sure if the first paragraph of my 'proof' implies the second.  I've shown that for arbitrary $h \in H$, $h \in H$ and possibly $h \in K$, and similar for $k \in K$.  I don't know how to wrap it up (or perhaps this route won't lead anywhere at all).  If this way won't work, I'd just like a hint on a new direction to take.  Thanks.
124218
Prove that if $k\mid m$, then $Z_m$ has a subgroup of order $k.$  Ok, so this doesn't look like too hard of a problem. So do I just show that it is closed under multiplication and inverse? I just don't know how I would do that with a problem like this. Can anyone help me get started please?
54470
Let $G$ be a simple group and there exists a subgroup $H$ of index $n \in \mathbb{N}_{\ge 3}$. Show that: $|G|$ divides $n!/2.$ I saw a similar question .   There was written: "The proof is really simple. Let $X$ be the set of left cosets of $H$. Consider $\phi: G \to \text{Sym}(X)$ given by $\phi(x)(aH)=(xa)H$. Then $\phi$ is a homomorphism. Consider now $K=\ker \phi$. Then $K$ is a normal subgroup of $G$ contained in $H$. Finally, $G/K$ is isomorphic to a subgroup of $\text{Sym}(X)$, which has order $n!$, where $n=[G:H]$. Thus, $[G:K]$ is finite and divides $n!$."  In fact, I don't understand the last sentence. "Thus, $[G:K]$ is finite and divides $n!$." Also did I understand it right that $ker(\phi)=H$?  Can you also give me an advice how to proof it for $n \ge 3$. (maybe induction)  Thank you for taking your time.
26739
How to show if $H,K$ are normal subgroups in a group $G$ then $[H,K] \leq H \cap K$.  Here $[H,K]$ is the commutator subgroup.  I have tried to start it but instantly failed:   I want to show firstly that $[H,K]\subseteq H \cap K$ but I don't understand what an arbitrary element of $[H,K]$ would even look like so I have something to work on.  I thought it could look something like $h^{-1}k^{-1}hk$ but then I thought that we are working with the smallest subgroup containing elements of this form but that might not be what the actual elements are?  Could anyone help?
18570
Note K is a normal subgroup of HK  I thought I could just prove K is a subset of HK and vice versa so as to prove it a normal subgroup but then how to do this I have no idea because I keep getting hk element again while trying to do so.  I just tried to implement the steps to find HK is a subgroup of G.
2057
Respected all  Please help me on this problem.   Suppose that $G$ is given group and $N$ be a normal subgroup of $G$ and $N$ is isomorphic to $G_1$ as groups. If $G/N$ is isomorphic to the group $G_2$, can we establish that $G$ is isomorphic to the group $G_1\times G_2$ ?
109600
I was having a discussion with a friend of mine about some normal group properties and then came up with the question "if G is not commutative, then is there always a subgroup that is not a normal subgroup?" It's probably more easy to solve this in the following form:$$\forall H \leq G : H  \lhd G \Rightarrow \forall a,b \in G : ab=ba$$ My question is, can anybody give a proof, or a counter-example (because I don't think it holds) of this theorem? Thanks!
77409
The following is a problem from the Berkeley Problems book.      Let $G$ be a group with three normal subgroups $N_1$ , $N_2$, and   $N_3$. Suppose $N_i \cap N_j = \{ e\}$ and $N_iN_j  = G$ for all   $(i,j )$ with $i \ne j$ . Show that $G$ is abelian and $N_i$, is   isomorphic to $N_j$ for all $i, j$ .   Attempt:   I have not got very far. The only thing that comes to my mind is that as $|HK|=\frac{|H||K|}{|H \cap K|}$, $|N_1|=|N_2|=|N_3|=m$, and order of $|G|=m^2$. I have made other futile attempts of trying to show m prime, finding an isomorphism between the $N's$. I think the fact that the equation $N_i N_j=G$, could be seen as, left coset of $N_j$ with the multiplying elements restricted to only $N_i$ could provide me an answer, but I haven't got anywhere with this.   Thanks in advance.
32431
Let $G$ be a group and $H$ be a normal subgroup of index $p$ ( a prime ) ; suppose $K$ is a subgroup of $G$ not contained in $H$ , then is it true that $G=HK$ ? ( I know that the fact is true if $p=2$ , but I don't know for the general case . Please help . Thanks in advance )