# Tag Info

5

It is a finite group, but its order is not always $pn$. If $p$ is even then the order is $pn$, whereas if $p$ is odd, then the order is $p$ when $n$ is odd and $2p$ when $n$ is even. There is no need to assume that $n \ge p$.

5

You are misstating Cayley's theorem. It says that if $|G|=n$, then $G$ is isomorphic to a subgroup of $S_n$. Now, when $\phi:G\to H$ is a homomorphism and $G$ is abelian, it follows that for $a,b\in G$, $$\phi(a)\phi(b)=\phi(ab)=\phi(ba)=\phi(b)\phi(a)$$ so $\phi(G)$ is an abelian subgroup of $H$. For example, the cyclic group $C_n$ embeds as the cyclic ...

4

For a counterexample to question 1, compare the dihedral group $D_4$ (of order 8) with $\mathbb Z_4\times\mathbb Z_2$. Both groups are generated by two elements $a,b$, with the common relations $a^4=b^2=1$, and the additional relation $bab=a^{\pm 1}$, with exponent $+1$ for the abelian group and $-1$ in the case of $D_4$. The elements are $a^n$ and $a^nb$, ...

4

The way this question is usually worded is "Prove if $H$ is a subgroup of $G$ with index $2$, then $H$ is normal in $G$." If you're just looking for an answer, you can find this proof easily with a google search. However if you want to give this a go yourself here's a hint: We have two left cosets whose disjoint union is the whole group: $H$ and $xH$ for ...

3

If $G/\Phi(G)$ is not simple, then let $N/\Phi(G)$ be a minimal normal subgroup. Then $N < G$, so $N$ is solvable. Let $M$ be a maximal subgroup of $G$ not containing $N$ (such an $M$ must exist since $N$ is not contained in $\Phi(G)$). Then $MN=G$, and since $N$ is solvable but $G$ is not, $M$ cannot be solvable, contradicting $G$ being minimal ...

3

The answer appears to be no. A nilpotent group in which every $2$-generator subgroup has class $2$ has class at most $3$. There is an example of order $3^7$ of such a group of class $3$. However, for $n \ge 3$, if all $n$-generator subgroups of $G$ have class $n$, then so does $G$. These results are apparently due to Levi and Heineken. See here.

3

I believe you have the wrong copy of $\mathbb{C}$ inside $\mathbb{C}[G]$. Instead, you want to take $\mathbb{C}\cong\{c\cdot\delta_e\mid c\in\mathbb{C}\}$, where $\delta_e$ is the function $$\delta_e(g)=\begin{cases}1&\mbox{if }g=e\\0&\mbox{otherwise.}\end{cases}$$ In fact, there is an isomorphism of algebras ...

3

HINT: $$G_1 \le G_1 \times G_2 \le G_1 \times G_2 \times G_3 \le \dots \le G_1 \times \cdots \times G_n$$ is a composition series for $G$. You can see this noting that $G_1 \times \cdots \times G_{k-1}$ is an epimorphic image of $G_1 \times \cdots \times G_k$ simply by the projection, and the kernel is $1 \times \cdots \times 1 \times G_k \cong G_k$. Now, ...

3

Every finite group is the homeomorphism group of some finite $T_0$ space. This follows fairly easily from the fact that a $T_0$ topology on a finite set is the same thing as a partial order. Here is a simple construction of a poset with a given automorphism group; there are many others you could give. Let $G$ be any nontrivial finite group and let ...

2

The linear characters are precisely the characters (in the trace sense) of $1$-dimensional representations, which are automatically irreducible.

2

Hint: The size of any orbit under the action of conjugation divides the order of $G$ by the orbit stabilizer theorem, so the size of every conjugacy class (orbit) is odd. Since the orbits (conjugacy classes) partition $G$, what happens if there are an even number of conjugacy classes?

2

Every group $G$ of order $n$ such that $(n,\phi(n))=1$ is in fact cyclic, hence abelian. For an elementary proof see one of the following references (the first proof is simple enough to be suitable for an elemenatry class in group theory, the author says): Jungnickel, Dieter. On the Uniqueness of the Cyclic Group of Order $n$. Amer. Math. Monthly, Vol. 99, ...

2

In answering this problem I will assume that we are allowed to use a computer algebra system since it is somewhat computation-intensive but not much is gained by doing these computations with pen and paper. Observe that the convention at the OEIS is that the term necklace refers to the slots being arranged around a circle with rotational symmetry ...

2

In general, an action of a group $G$ on a set $X$ is equivalent to a homomorphism $\varphi: G \to \text{Sym}(X)$, where $\text{Sym}(X)$ is the set of all permutations of $X$, i.e., bijections $X \to X$. (This is called a permutation representation; see here for more.) In this problem, the set $N$ (on which $G$ acts) has the structure of a group, and the ...

2

This can be done more generally. For simplicity, we denote by $E_{12}$ the element $$\begin{bmatrix} 1 & 1 & & & \\ & 1 & & & \\ & & 1& & \\ & & & \ddots & \\ & & & & 1\\ \end{bmatrix}$$ here the other entries are $0$. In general, then let $E_{ij}$ denote ...

2

What you say cannot be proved. Suppose you have that $n$ is coprime with $G$. Then you can consider the cyclic group $G'=G \oplus C_n$ (where $C_n$ is the cyclic group of order $n$). Clearly you have the projection $$p:G' \longrightarrow G$$ so now consider $\phi_1'=\phi_1 \circ p$ and $\phi_2'=\phi_2 \circ p$. $G',\phi_1', \phi_2'$ satisfy the same ...

2

You have an action of $\mathbb Z_5$ on the set of all colorings of the windmill (a set with $\binom{10}{4,3,3}$ elements). By Burnside the number of orbits is: $$\frac{1}{|G|}\sum\limits_{g\in G}X^g.$$ There are only two types of permutations in play. The identity clearly leaves all $\binom{10}{3,3,4}$ colorings unchanged. Any other rotation: This ...

2

Let me typeset some of the content from the comments. We have five panels attached to the center and five is prime, so the cycle index is really simple here namely $$Z(C_5) = \frac{1}{5} (a_1^5 + 4 a_5).$$ Taking into account that the outer and inner elements of the panels move in sync but never enter into the same orbit we get $$Z(W) = \frac{1}{5} (a_1^{10} ... 2 A (cyclic) group of order 3 is always of the form$$\{e,a,a^{-1}\} $$where e is the identity, and a and a^{-1} have order 3. Thus the number 10 is the result of the calculation$$\frac{20}{3-1}=\frac{20}{2}$$(the identity element is not of order 3). Similarly$$6=\frac{24}{5-1}. $$1 As Derek Holt points out, S_{n+2} with the natural action works. But there are smaller groups as well, even under the constraint that the action is transitive; for example, take (\mathbf{Z}/n\mathbf{Z})\times S_3 acting diagonally on [n]\times[3]; (e,(12)) has n fixed points and (r,*) has no fixed points if r\ne e. 1 The subgroup generated by b is acting on the subgroup generated by g via conjugation (this works because \langle g\rangle is normal in G). The reason why the i = 2 and i = 3 case are "the same" is because b \mapsto b^3 is an automorphism of \langle b\rangle, whereas b \mapsto b^2 is not (b^2 = (b^3)^2, but b \neq b^3). The automorphism ... 1 The order of 11 is the smallest number k for which$$\underbrace{11 + 11 + \ldots + 11}_{k \textrm{ times}} \equiv 0 \pmod {26}.$$In other words, we have 11k \equiv 0 \pmod {26}, which is to say that 11k is a multiple of 26. But 11 and 26 are relatively prime, so k itself must be 26. Note that 12 is the order of 11 in ... 1 We have |H|=4, |N|=8, with H \lhd N. So any element g \in N \setminus H satisfies H\langle g \rangle = \langle g \rangle H. If h is any element of S_4 of order 3, then |\langle h \rangle| =3, so N\langle h \rangle and \langle h \rangle N must have order 24 and hence they are the whole of S_4, so they are equal. 1 Because the trace is an additive invariant of a representation. If V= \bigoplus a_iV_i is a G-module, then for each g\in G$$\chi_V(g) = Tr(\pi(g)) = Tr(\sum_i a_i\pi_i(g)) = \sum a_i Tr(\pi_i(g)) = \sum a_i\chi_i(g). $$In view of the comments below, let me supplant my answer with the rest of the story (I am working over \mathbb{C} from now on). ... 1 We know that for p-groups, the length of upper and lower central series is same. Thus, if |G|=p^n, n\geq 3, and if G is of maximal class then G has upper central series$$1 < Z_1(G) <Z_2(G) <\cdots < Z_{n-2}(G) < G$$where |Z_i(G)|=p^{i}. Since, the lower central series also has same length, it is$$G >\gamma_2(G) ...

1

Every group $G$ of even order has an element of order $2$. The trick is to pair each element of $G$ with its inverse; the elements which are not of order $\leq 2$ come in pairs, so the number elements which are of order $\leq 2$ must be even. Since the identity has order $1$, the number of elements which are of order $2$ must be $\geq 1$. (I can explain this ...

1

Hint: $2^n\equiv -1$ mod $(2^n+1)$.

1

Hint: the order of $A_{4}/V_{4}$ is $12/4 = 3$. How many groups of order $3$ are there up to isomorphism? (Feel free to comment if you need more).

1

Note $S_{4}$ is a group of order $24$, not $4$. In fact, there are only two groups of order $4$ up to isomorphism: $\mathbb{Z}_{4}$, the cyclic group of order $4$, and $V_{4} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{2}$, the so-called Klein 4-group. $\mathbb{Z}_{4}$ is distinguished from $V_{4}$ by the fact that it has an element of order $4$, whereas $V_{4}$ ...

1

Note that any homomorphism of $\mathbb Z_5$ is determined by the image of a generator. It is an automorphism if that image is non-trivial. To be a little more explicit about things: If $y$ is a generator, $\alpha$ is an automorphism and $\alpha(y)= y^2$ then $\alpha (y^r)=y^{2r}$ If we apply $\alpha$ twice, $\alpha^2(y)=y^4$, three times gives ...

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