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8

There is no group object: since you only have bijections in your category, for all G there exists an X such that hom(x,G) is empty, so not a group in any way at all.


8

It is true: By taking repeated powers $x, x^2, x^3, ...$ of an element $x \in H$, we eventually must have a repeat $x = x^n$ for some $n>1$ (why?). This implies that $x^{n - 1} = e$ and that $x^{-1} = x^{n - 2}$, so $H$ is closed under inverses as well as products.


6

Complementary set of primes: all the other primes. Concretely, if $\mathscr P=\{$all primes$\}$, then $$\pi'=\mathscr P\setminus\pi,\quad p'=\mathscr P\setminus\{p\}\,.$$ A $p'$-group consists of elements that all avoid the prime $p$ in their orders.


6

${\rm PSU}(3,3)$ of order $6048$ is an example in which this is not possible. Holt and Rowley showed that this is the smallest simple example.


3

Since \begin{align} \varphi(14) &=14\left(1-\frac{1}{2}\right)\left(1-\frac{1}{7}\right)=6 \\[6px] \varphi(18) &=18\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)=6 \end{align} we know the two groups can be isomorphic. If we consider $3\in U_{14}$ and its powers, we get $$ 1=3^0,\quad 3=3^1,\quad 9=3^2,\quad 13=-1=3^3 $$ so the order of $3$ is ...


3

You can't necessarily say that it is $30$, but for a finite group $G$, the order of an element divides the order of the group. Thus, you can say that $6||G|$ and $10||G|$, so...?


2

No, in general it's not the case. Morally, I think it can be seen from the fact that your equation only involves product in the second argument of $\rho$ which creates a dissymmetry. Concretely, consider $G = \mu_3 = \{1,j,j^2\}\subset U(1)$. I define $\rho: G\times G\to U(1)$ by $\rho(1,g)=\rho(g,1)=1$ for all $g\in G$, $\rho(j,j^2)=j$, $\rho(j^2,j)=j^2$, ...


2

The following algorithm will produce $H=\langle S\rangle$, provided $S$ is a subset of a finite group $G$. Push $1$ to a queue and let $H=\emptyset$. If the queue is empty, terminate Pop $x$ from the queue. If $x\in H$, go back to step 2 Set $H=H\cup \{x\}$ For each $s\in S$, push $x\cdot s$ to the queue Go back to step 2


2

There is a very simple proof that only requires $H$ to be finite (and non empty). For $x\in H$, consider the map $f_x\colon H\to H$ defined by $f_x(h)=x*h$. This map is injective (prove it), so it is surjective because $H$ is finite. Then there exists $h_0\in H$ such that $$x=f_x(h_0)=x*h_0=x;$$ therefore $h_0=e$ and so $e\in H$. But then we have also ...


2

I've calculated this for $1 \le n \le 10^5$ using ParListWithSCSCP (see here) with one master and $48$ workers. This is an embarrassingly parallel calculation, and it took about half an hour on a 64-core Linux server with AMD Opteron 6375 1400MHz CPUs. It confirms that the next entry is given by $gnu(81900)=4087$, and there is no further champion for $81900 ...


2

You need a map $\text {Aut} (G) \times G\to G$. Define it by $(\phi, g) \mapsto \phi (g)$. To show that this is an action you need to check that $(\phi_1\phi_2,g)$ and $(\phi_1,(\phi_2,g))$ map to the same element and that $(\text {Id}, g) \mapsto g$


2

This is not complete answer; but a partial information, which in addition to Holt's comment may simplify your job. (I will try to write complete proof as I get some directions on it) Suppose $G$ is a group of order $2^5$ satisfying conditions in your question. We show that $Z(G)=C_2\times C_2$. For this, by hypothesis, $|Z(G)|=2^2$ or $2^3$. If ...


2

Lagrange is the way to go. Note that $H\cap K$ is a subgroup of $H$ and $K$, so its order must divide $|H|$ as well as $|K|$...


2

The answer is yes, although all the proofs use the classification of finite simple groups. Your question was a conjecture of Brauer, which was proven in the following paper (theorem 3.6 in there). Kimmerle, Wolfgang; Lyons, Richard; Sandling, Robert; Teague, David N. Composition factors from the group ring and Artin's theorem on orders of simple ...


2

Consider $n_p$, the number of $p$-Sylow subgroups. $n_p$ satisfies $n_p\mid 12$ and $n_p\equiv 1\pmod p$. If $p=7$ or $p>11$, the only possibility of $n_p$ is 1, so $G$ has a non-trivial normal subgroup. If $p=11$, $n_p$ is either 1 or 12, $n_2=1$ or $n_2\ge 3$ and $n_3$ is either 1 or greater than or equal to 4. We are going to rule out the case ...


1

You've misread the notation. (12) is not an element of H - (12)(34) is. It sends 1234 to 2143. Then (14)(23) sends that to 3412, which is also (13)(24).


1

I think that ${\mathtt{SmallGroup}}(300,25)$ is a counterexample. It has Fitting length $3$, with $|F_1|=25$, $|F_2/F_1|=6$ and $|F_3/F_2|=2$. It has two normal subgroups of index $2$ with Fitting length $3$, but they are interchanged by an automorphism of $G$, so they are not characteristic. $F_1$ is the unique minimal normal subgroup, so all proper ...


1

Another definition of a group $\Gamma$ acting on a set $\Omega$ is a group homomorphism $\Gamma \to S(\Omega)$, where $S(\Omega)$ is the group of bijections of $\Omega$. With this definition, and the fact that every automorphism of $G$ is a bijection of $G$, we get that inclusion is an (injective) homomorphism $\Gamma = Aut(G) \to S(G)=S(\Omega)$.


1

If $r$ is odd, then $x^r=1$ implies $x=1$. Thus the kernel is trivial, thus the map is injective.


1

Let $g\in N_G(H)$. Then $g^{-1}Hg=H$. Thus $Hg=gH$ and thus $H$ is a normal subgroup of its normalizer. Let $h\in H$. Certainly $h^{-1}Hh=H$ due to the closure property of the subgroup $H$. Thus $h\in N_G(H)$. This shows that $H$ is contained in $N_G(H)$.


1

Yes, there are two different $A$'s and two different $\chi$'s, which is confusing. You at least resolved the problem of the $A$'s correctly. $A,B$ in the direct product need not be cyclic (in fact, we may not be able to make both cyclic). What we need is that for every abelian non-cyclic $\Bbb A$, there is a product decomposition as $A\times B$ with smaller ...


1

The author is assuming that we have an abelian group $G$ that decomposes as $$G= A \times B,$$ where $A$ and $B$ are cyclic groups. Furthermore, we get to assume that for all characters $\chi$ we have $\chi((a_0,b_0))=1$, for some $(a_0,b_0) \in A \times B$. But since the map $$\varphi_{\chi}: A \times B \to T$$ given by $\varphi_{\chi}((a,b))=\chi(a)$ for ...


1

$(2,0)+(2,0)=(0,0)$ which is the identity. Thus it's order is 2. Another example: the order of $(1,1)$ is 4.



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