Let H be a subgroup of S4 where $H = \{e, B , C ,D \}$  $B(1)=2,B(2)=1,B(3)=4,B(4)=3$  $C(1)=3,C(2)=4,C(3)=1,C(4)=2$  $D(1)=4,D(2)=3,D(3)=2,D(4)=1$  Prove that H is a normal subgroup.  I've tried using the definition that $H$ is normal $\iff gH=Hg$, $\forall g$  As I can't take all the 24 elements of S4 and check it,I've tried with a general form: $g(1)=a,g(2)=b,g(3)=c,g(4)=d$. I managed to find $gH$,but $Hg$ isn't as easy.  What I've also noticed is that $BC=D,CD=B,DB=C, B=B^{-1} , C=C^{-1} , D=D^{-1}$ and $B^{2}=C^{2}=D^{2}=e$.   Can you give me a hand, please?
34810
Can anyone tell me how to find all normal subgroups of the symmetric group $S_4$?  In particular are $H=\{e,(1 2)(3 4)\}$ and $K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}$ normal subgroups?
27938
Let $G$ be a group, and $H$ a subgroup.  $H$'s normalizer is defined: $N(H):=\{g\in G| gHg^{-1}=H \}$.   Prove $N(H)$ is a normal subgroup of G, or give counterexample.  Intuitively it seems to me that this claim is wrong, however, I'm having trouble with finding a counterexample.  Thans in advance for any assistance!
57433
I'm trying to understand what the relation is between the direct product and the quotient group.   If we let $H$ be a normal subgroup of a group $G$, then it is not too difficult to show that the set of all cosets of $H$ in $G$ forms a quotient group $G/H$: \begin{equation} G/H = \{ g H \mid g \in G \} \end{equation}  On the other hand, the Cartesian product of two groups $G$ and $H$ is defined as: \begin{equation} G \times H = \{ (g,h) \mid g \in G \text{ and } h \in H \} \end{equation} where $(g,h)$ denotes the set of ordered pairs. The direct product operation on this set is defined as: \begin{equation} (g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2) \in G \times H \end{equation} and it is easy to see that the direct product forms a group.  Is the following statement true: \begin{equation} K = G \times H \implies G \simeq K / (\{e_G \} \times H) \end{equation} If so, under what conditions is it true? And how can we see it is true (or false)?  Edit 26/03:  Up to this point, I believe I have found a method (see below) of showing the isomorphism relation. I would be really grateful if someone could tell me whether this proof is correct or not.  Let us identity the elements of $h \in H$ with element of $K$ by setting $h \equiv (e_G,h)$. The elements of $K/H$ are as usual defined by: \begin{equation} K/H = \{ (g,h) H \mid g \in G \text{ and } h \in H \} \end{equation} Since $h_1H=H$ for some $h_1 \in H$, we have: \begin{equation} (g,h)H = (g',h')H \iff g=g' \text{ and } h' = h h_1 \tag{1} \end{equation} and so without loss of generality we can write every element of $K/H$ in the form $(g,e_H)H$. Now, let the map: \begin{equation} f : G \to K/H \end{equation} be defined by: \begin{equation} f(g) = (g,e_H) H \tag{2} \end{equation} The map is one-to-one. This can be seen by equation $(1)$, because if: \begin{equation} f(g) = f(g') \end{equation} then: \begin{equation} (g,e_H) H = (g',e_H) H \implies g=g' \end{equation} Furthermore, the map is trivially onto: \begin{equation} \forall (g,h) H \in K/H \; \exists \; g \in G \; , \; f(g)=(g,h) H \end{equation} and thus the map is bijective. Finally, the map is also a homomorphism, because: \begin{equation} f(gg') = (gg',e_H) H = (g,e_H)(g',e_H) H = (g,e_H)(g',e_H) HH = (g,e_H) H (g',e_H) H = f(g) f(g') \end{equation} and so $f$ is a isomorphism. Thus, by definition of equation $(2)$, we have shown that $G  \simeq K/H$.  Any input is much appreciated.
5791
I'm looking for verification:  My claim: If $G$ is a finite group and $H$ is a (proper)subgroup of index $k&gt;1$, where $k! \leq |G|$, then $G$ is not simple.  Proof: Consider the set of left cosets of $G$ by $H$:  $$G/H = \{H,g_1H,\dots,g_kH\}$$  Then $\phi: G \rightarrow S_{G/H}$ is a homomorphism where we define:  $$\phi(g)[g_iH] = g g_iH$$  ($S_{G/H}$ is the group of permutations on elements of $G/H$ and of course $S_{G/H} \cong S_k$.)  Also $\ker \phi \neq G$ because $\phi(g_1)$ is clearly not the identity in $S_{G/H}$.  Because $\phi$ is a homomorphism,  $$ G/\ker\phi \cong \text{im}\phi$$  Then $|\text{im} \phi| \leq |S_k| = k!$ and if $k! \leq |G|$ then $|\ker\phi| &gt; 1$ and $\ker\phi$ is a non-trivial normal subgroup of $G$.  Example: If a group of order $36$ has a subgroup of index $3$ or $4$ then the group is not simple.
37197
Let $H$ be a subgroup of a group $G$ such that $x^2 \in H$, $\forall x\in G$. Prove that $H$ is a normal subgroup of $G$.    I have tried to using the definition but failed. Can someone help me please.
114826
How to find the number of subgroups of $S_4$ and list them?   For order $1$ it is {${e}$}, order $2$, we will consider   $&lt;(12)&gt;$, $&lt;(23)&gt;$, $&lt;(34)&gt;$, $&lt;(13)&gt;$, $&lt;(14)&gt;$, $&lt;(24)&gt;$, for order $3$ we will consider $&lt;(123)&gt;$, $&lt;(124)&gt;$, $&lt;(143)&gt;$, $&lt;(234)&gt;$, for order 4 we will look at $&lt;(1234)&gt;$, $&lt;(1324)&gt;$, and so on. $A_4$ will also be there. I don't think these will be all ...
104817
Let $G$ be a finite group and let $H$ be a subgroup of $G$ with the property that no other subgroup of $G$ has order $|H|$. Prove that $H$ is normal.  My attempt:   To show that $H$ is normal, I need to show that for any $g,h \in G$, $ghg^{-1}\in H$.  Let the order of $G=n$ and let the order of $H=k$. If $H$ is a subgroup of $G$ then $k$ divides $n$ and the identity element $e$ is both $G$ and $H$. Also for any $a,b \in H$, $ab^{-1}\in H$.  Since $k$ divides $n$, $n=qk$ for some integer $q$.  $a^{n}=a^{qk}=e$...  (I don't think this proof is getting me anywhere. I think I need a hint about what does "$H$ being a subgroup of $G$ such that no other subgroup of $G$ has order $|H|$" imply...)
155
Question: Let G be a group and let G' be the subgroup of G generated by the set $S=\left \{ x^{-1}y^{-1}xy \mid x,y \in G \right \}$  $\space$  Prove that G' is normal in G. Solved  $\space$  Prove that G/G' is abelian. Solved  $\space$  Prove that if H is a subgroup of G and G' is a subgroup of H, then H is a normal subgroup of G.  I cannot even begin to start on this question. Any hint is appreciated.
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?
96709
Let $G$ be a group and let $f:G\to G$ be an isomorphism. Let $H = \{ a \in G \; | \; f(a)=a^{-1}\}$ Prove that $H$ is a subgroup of $G$ if and only if $G$ is abelian.  My attempt: I already know that $H$ is a subset of $G$ since $a$ is in $G$. And that $H$ is $a$ group if and only if for every $a$ and $b$ in $H$, $ab^-1$ is in $H$. That is, if $f(a)=a^{-1}$ and $f(b)=b^{-1}$ then $f(ab^-1)=(ab^{-1})^{-1}=(b^{-1})a$.  But I don't know how to derive this nor the converse based on the abelian property of $G$ that for every $a$ and $b$ in $G$, $ab=ba$.  Please help some. Thanks!
44531
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  Prove that H = ($\begin{pmatrix} 1 &amp; t \\ 0 &amp; 1 \end{pmatrix}: t\in\mathbb{R}$) is a normal subgroup of G  I'm meant to show that H is a normal subgroup by showing that the left coset is equal to the right coset.  So far I have shown the following:  take $g=\begin{pmatrix} a &amp; b \\ 0 &amp; 1 \end{pmatrix}$   then  $gH=\begin {pmatrix}a &amp;at+b \\ 0 &amp;1 \end{pmatrix}$  and take $g'=\begin{pmatrix} a' &amp; b' \\ 0 &amp; 1 \end{pmatrix}$  then $Hg'=\begin{pmatrix} a' &amp; b'+t \\ 0 &amp; 1 \end{pmatrix}$  I don't quite understand how $gH=Hg'$
110998
I'm trying to understand the proof that any group of index $2$ is normal.   Let $G$ be a group and $H$ a subgroup of index $2$.  I understand that if $x\in H$ then $xH=H=Hx$ since $H$ is a subgroup. However, I don't understand why if $x\in G\setminus H$ then $xH=G\setminus H.$  I know there are other proofs of this on this site but none of them specifically address this question.
3691
Please excuse the selfishness of the following question:     Let $G$ be a group and $H \le G$ such that $|G:H|=2$. Show that $H$ is normal.   Proof:  Because $|G:H|=2$, $G = H \cup aH$ for some $a \in G \setminus H$.   Let $x\in G$. Then $x \in H$ or $x \in aH$.  Suppose $x \in H$. Then $xhx^{-1} \in H$ because $H$ is a group.  Suppose $x \in aH$. Then $ahH(ah)^{-1} = ahHh^{-1}a^{-1} = a(hHh^{-1})a^{-1}$ ***1  ***1: Can I say now that $x \in H$, based on the that $hHh^{-1} \in H$ and that $aa^{-1} = e$ ?  Thank you for the time you spent reading my doubts.
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.
57017
Let $G$ be a finite group and $H$ a subgroup of index $p$, where $p$ is a prime. If $\operatorname{gcd}(|H|, p-1)=1$, then $H$ must be normal. Does somebody have a quick proof of this?
36159
Is there a better (more direct or intuitive) proof for this proposition than I have come up with below? I am not sure whether it could be simiplified:  Let $G$ be a group with $H \leq G$. Then $K = \bigcap_{g \in G} gHg^{-1}$ is normal in $G$.  Let $a \in K$. Then $a \in gHg^{-1}$ for all $g \in G$. Therefore for all $g_1,g \in G$, $g_1ag_1^{-1} \in g_1gHg^{-1}g_1^{-1} = (g_1g)H(g_1g)^{-1}$ and so $g_1ag_1^{-1} \in K$ since $g_1g \in G$. Then $K$ is normal in $G$.
70190
With the next group:  gap&gt; G:=Group((1,2,3,4),(1,2));  Group([ (1,2,3,4), (1,2) ])  gap&gt; Order(G);  24  gap&gt; Elements(G);  [ (), (3,4), (2,3), (2,3,4), (2,4,3), (2,4), (1,2), (1,2)(3,4),   (1,2,3), (1,2,3,4), (1,2,4,3), (1,2,4), (1,3,2), (1,3,4,2),   (1,3), (1,3,4), (1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2),   (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3) ]   Is there any function to list the subgroups of G? Thanks, and sorry if the question looks so easy, i'm a newbie in GAP.