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4

There are indeed $\frac156!=144$ different $5$-cycles in $S_6$ but note that a single $5$-Sylow accounts for 4 of them. Indeed $c$, $c^2$, $c^3$, $c^4$ are the 4 different non-trivial elements in $\langle c\rangle$. Thus the total number of $5$-Sylows is $\frac14144=36$.


3

I don't know whether it is true for almost simple groups, but it is not true in general. For $\pi = \{2,3\}$, ${\rm PSL}(2,7)$ has two non-conjugate Hall $\pi$-subgroups of order $24$, and we can use that to construct a counterexample. Let $G = {\rm PSL}(2,7) \wr C_5$ be the wreath product of ${\rm PSL}(2,7)$ and a cyclic group of order $5$. Then $G$ has a ...


2

It is true always and in any case that $\;H\le N_G(H)\;$ , for any subgroup $\;H\le G\;$ , now: (1) Proof with nilpotent groups: since $\;U\;$ is a $\;p\,-$ subgroup which is not a Sylow $\;p\,-$ subgroup, $\;U\;$ is properly contained in some Sylow $\;p\,-$ subgroup $\;P\le G\;$ , and $\;P\;$ is a nilpotent group as it is a finite $\;p\,-\;$ subgroup, so ...


2

There are so many easy and direct ways to see this is not true... Here is another one: Every subgroup of a cyclic group is cyclic. However, by the theorem you mentioned, $\mathbb Z_4 \times \mathbb Z_{12}$ is not cyclic, and it is (trivially isomorphic to) a subgroup of your group.


2

Claim. The elements of $G$ can be written as $\{e,a,b,ab\}$. Proof. Since $G$ is a group the identity element $e\in G$. Since $G$ is a group of order $4$, there must exist another three non-identity elements which are also mutually distinct. Let them be $a,b$ and $c$. We claim that $ab=c$. If not then $e,a,b,ab$ and $c$ all are elements of $G$. ...


1

Let $N$ be the set of non-generators. The set $G \setminus N$ is generating and has the maximum cardinality among all generating sets that do not contain an element of $N$, and it is unique.


1

Using the homomorphism property, $$f(2)=f(1+1)=f(1)^2.$$ What is $f(1)^2$ in the Klein 4 group? (Try out all the different possible values for $f(1)$ if you're not sure.)


1

No. As examples mentioned in the commentary, a subgroup of order p of a cyclic group of order $p^n, n>1$ is not a Hall subgroup, but it satisfies the property P.


1

Approach one: In this approach we directly show it is surjective. Given $(x,y)\in U(s)\times U(t)$. Consider $\phi(xt T+ys S)$ where $tT\equiv1~mod~(s),~sS\equiv1~mod~(t)$. Why does such $S,~T$ exists? Why is $xt T+ys S$ relative prime to $st$? What is $\phi(xt T+ys S)$? Approach two: In this approach we assume we know the fact that Euler phi function ...


1

You can easily show the group contains two normal subgroups of size $2$. There is a criterion that states that if $K$ and $H$ are normal subgroups of $G$ with $K\cap H =\{e\}$ and $KH=G$ then $G\cong K \times H$.


1

Let it be that $[G,G]\leq H\leq G$. We have: $$ghg^{-1}=ghg^{-1}h^{-1}h=ah$$ where: $$a:=ghg^{-1}h^{-1}\in[G,G]\leq H$$ So $h\in H$ will imply that $ghg^{-1}=ah\in H$. This shows that $H$ is normal. $G/H$ is abelian since $b^{-1}a^{-1}ba\in H$ (which is true for each pair $a,b\in G$) implies that $abH=baH$ or equivalently $(aH)(bH)=(bH)(aH)$.


1

Yes that arrow is bad. You can use $\odot$ and $\otimes$ to denote arrow against the viewer or arrow forwards. $\odot$ being the tip of a dart facing you (arrow pointing your direction) $\otimes$ being you seeing the feathers of the dart (vector) as you throw it forward from your p.o.v.



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