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9

Let $G = S_5$, the symmetric group of 5 elements. Let $x_1 = (1,2), x_2 = (3, 4), x_3 = (1, 5), x_4=(2, 4), x_5 = (3, 5).$


8

It's the Frobenius group $C_5\rtimes C_4$. Hint: Start with $$\sigma:\left\{\begin{array}{l}2^{1/5}\mapsto \rho 2^{1/5}\\ \rho\mapsto \rho\end{array}\right.\hspace{15pt}\text{and}\hspace{15pt}\tau:\left\{\begin{array}{l}2^{1/5}\mapsto 2^{1/5}\\ \rho \mapsto \rho^2\end{array}\right.$$ To generalize, take a look at the order $\bmod p$ of the power in $\tau$, ...


5

By Sylow's Theorem, we have that $P_5\trianglelefteq G$ is normal, which we expect since $\Bbb Q(\rho)/\Bbb Q$ is Galois. Since we know that $\text{Gal}(\Bbb Q(\rho)/\Bbb Q)\cong \Bbb Z/4$ we see that there is a surjection $G\to\Bbb Z/4$, hence we see that we have the short, exact sequence: $$1\to \Bbb Z/5\stackrel{i}{\longrightarrow} ...


5

No. Consider elementary matrices. Let $e_{ij}$ denote the matrix with $(i,j)$ entry $\delta_{ij}$. Let $e_{ij}(\lambda)=1+\lambda e_{ij},i\neq j$. Assume $n>2$. We have the so called Steinberg relations $$\begin{align}&(1)& e_{ij}(\nu)e_{ij}(\mu)&=e_{ij}(\nu+\mu)\\ &(2)& [e_{ij}(\nu),e_{jk}(\mu)]&=e_{ik}(\nu\mu)&\text{ if ...


4

If $A$ is a subgroup of ${\rm Aut(G)}$ with $C_G(A)=1$, then one says that $G$ acts fixed-point-freely upon $G$ by automorphisms. The fixed-point-free automorphism conjecture asserts that if a finite group $G$ admits a fixed-point-free automorphism group $A$ (and, if $A$ is noncyclic, further suppose that $gcd(|G|, |A|) = 1$), then G is soluble. Several ...


4

There exists an element $a$ of order $3$. If there was not every element would need to have order $11$. But then a counting argument leads to contradiction. We take the quotient group $ G/<a>$. Then it has order $33$. The same argument shows that there exists an element $b$ of order $3$. Take the map $p: G \rightarrow G /<a>$ the natural ...


3

The fact that there are more than $n/p^2$ irreducible characters whose squares sum to $|G|$ implies that at least one nontrivial character $\chi$ has degree less than $p$. Since $G$ is simple, $\chi$ is faithful, so it restricts to a faithful character $\chi_P$ of a Sylow $p$-subgroup $P$ of $G$. The degrees of the irreducible constituents of $\chi_P$ all ...


3

If $n_3=4$ you have $8$ elements of order $3$. so there are $4$ elements of other orders. Those are exactly enough to have a subgroup of order $4$, so it is unique. Hence normal.


3

I will write a very useful lemma, Lemma: If the index of $H$ in $G$ is the smallest prime dividing $G$, then $H$ is normal. This lemma proves $1$ directly. 2) Notice that whhen you show that it has a uniqe Sylow-p subgroup, you also shows that it is normal in G. $b)$ Let $H$ be the subgroup of order $7$ and $K$ be a subgroup of order $4$. It is clear ...


3

Just to belabor the obvious, the (unstated) assumption that the groups in question are finite is essential. For instance, the integers and the even integers are both subgroups of the rationals, with $[\mathbb Q : \mathbb Z] = [\mathbb Q : 2\mathbb Z] = \infty$, and $\mathbb Z \subset 2\mathbb Z$.


2

By Cayley's theorem, a group $G$ of order $30$ is isomorphic to a subgroup of $S_{30}$ is such a way that no non-identity element of $G$ has any fixed point. There is an element $t$ of order $2$ in $G$ which is represented by a product of $15$ $2$-cycles, so as an odd permutation. The elements of $G$ which are represented as even permutations of $S_{30}$ ...


2

1) Consider the action of $G$ on the lateral classes of the subgroup ( call it H ) $$ g \cdot kH = gkH $$ This is an action. Consider the kernel K of the action, i.e $$K = \lbrace g \in H | g \cdot wH = w H \ \ \forall \ wH \rbrace$$Then $K \trianglelefteq G$; moreover $K \subseteq H $ ( simply consider $wH = H $ in the definition of $K$). $G/K$ is ...


2

Another approach, using the rather important property that finite $\;p$- groups have non-trivial center, which also gives for free the existence of such groups (and even of a normal subgroup of order $\;p^k\;$ , for any $\;0\le k\le n\;$. Take $\;1\neq z\in Z(G)\implies \langle z\rangle\lhd G\;$ , so $$\left|G/\langle z\rangle\right|=p^{n-1}$$ Apply ...


2

Hint: Consider $$\frac{|G|}{|A|}=\frac{|G|}{|B|}\frac{|B|}{|A|}.$$ So $[G,A]=[G,B][B,A]$. Then, under your condition $[B,A]=1$.


2

Yes this holds for Fermat primes. You need the following observation. This is sometimes called the $N/C$ Theorem. Lemma Let $G$ be a group with a subgroup $H$, then $N_G(H)/C_G(H)$ embeds homomorphically into Aut$(H)$. Here $N_G(H)=\{g \in G : gH=Hg\}$, the normalizer of $H$ in $G$ and $C_G(H)=\{g \in G : gh=hg \text { for all } h\in H\}$, the centralizer ...


2

Theorem :(first sylow theorem) suppose that $G$ is a finite group and $|G|=p^{n}m$ such that $p \nmid m$ then 1.$G$ include a subgroup of order $p^i$ for $1\leq i \leq n$ 2. Any subgroup $H$ from $G$ from order $p^i$ is normal subgroup of a subgroup from order $p^i$ Answer : if $G$ be not cyclic then for every elemen except $e$ must have order $p$. Let ...


2

Let's think about $S(p, q)$ and $S_p \times S_q$. $S(p, q)$ has $\binom{p+q}{p}$ elements. Think of $n$ adjacent slots: we have to choose $p$ of them to put the first $p$ in order, and afterwards the $q$ remaining elements must go in the consecutive remaining slots. $S_p \times S_q$ has $p!q!$ many elements: $p!$ permutations of the first $p$ elements and ...


2

Here's how I would view the problem; I don't actually use Mackey's formula, but, in this special case, Frobenius reciprocity leads to the same thing. Inducing the trivial character of $H$ up to $G$ produces the permutation character of $G$ acting on a transitive $G$-set with $H$ as one of the point-stabilizers, i.e., the $G$-set $G/H$. The inner product you ...


2

Let $M$ denote the number of cyclic subgroups of $\mathbb{Z}_{p^2} \times \mathbb{Z}_{p^2}$ of order $p^2$. If I'm understanding correctly, it seems that you're really only asking why $M = p^2 + p$. First, you need to be aware that: 1) If $G_1, G_2$ are groups then $(a,b) \in G_1 \times G_2$ has order lcm$(|a|, |b|)$. 2) A cyclic group of order $p^2$ will ...


2

Hint: if $\chi \in Irr (G)$, and $\phi$ is an irreducible constituent of $\chi_H$ (so $[\chi_H,\phi] \geq 1$), then by Frobenius Reciprocity $\phi^G(1)=\phi(1)[G:H] \geq \cdots + \chi(1) + \cdots \geq n$.


2

$223$ is a prime number. By Sylow theorems, if $N$ is the number of $223$-sylow subgroups, then $$N \equiv 1 \mod (223) $$ $$N \mid 3^2$$ This implies that $N =1$. Thus there is an unique $223$-sylow , which is therefore normal. Suppose we call it H. Consider $G/H$, it is a group of order $9$, and the projection $$\phi : G \to G/H $$ is a surjective ...


2

The Quaternion group $Q_8$ with $\{1,-1\}$ If a group satisfies that it has to be of prime power order and the subgroup in question has to be of prime order.Since groups of order $p^2$ are abelian the next candidates would be of order $16$ and $27$.


1

One thing that you may have overlooked is that for any prime $q$ there is an enormous amount of non-abelian $q$-groups, so take one, call it $Q$, and all direct products $\mathbf C_{p^i} \times Q$ will qualify provides $p<q$. In fact this is just one instance of the following generalization of your construction. Let's say a group satisfies $\star_p$ if ...


1

Take any two different odd primes $\;p\,,\,q\;$ . s.t. $\;p>q\;,\;\;q\mid(p-1)\;$ . Take two cyclic groups $\;C_p=\langle\,y\,\rangle\,,\,\,C_q=\langle\,x\,\rangle\;$ of order $\;p\,,\,\,q\;$ , resp. Then you can build a(n exterior) semidirect product $\;C_q\rtimes C_p\;$ by means of the homomorphism $$\;f: C_q\to ...


1

I think you are right. The set $X$ is the set of all two element subsets of $G$. For the action of $G$ on $X$, first note that $G$ has a regular action on itself. Use this action to define an action on $X$. More precisely, if $g_i \mapsto h_i$ and $g_j \mapsto h_j$ under the regular action of $G$ on itself $(i \neq j)$, then the action of $G$ on $X$ is ...


1

The two element subsets: $$\{0,1\},\{0,2\},\{0,3\},\{0,4\},\{0,5\},\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\}.$$ Left addition: Take $\left[a\right]_6 \in \mathbb{Z}_6$ and do $a+H$ (where $H$ is a two element subset of $G$). For example: $$1+\{0,1\} =\{1,2\}, 1+\{1,2\} = \{2,3\}, 1+\{2,3\} = \{3,4\}, 1+\{3,4\} = ...


1

If you can show that $G$ has an element $g$ of order 3, then $\langle g \rangle$ is a normal subgroup (since $G$ is abelian), and therefore $G / \langle g\rangle$ is an abelian group. Since the order of $G / \langle g\rangle$ is 33, this group also contains an element, say $x + \langle g\rangle$, of order 3. If $n$ is the order of $x$ in $G$, then $(x + ...


1

Direct counting of elements can handle this without even using factor groups (although to understand group theory better, it's good to understand the answers here based on factor groups). There is one element of order $1$. This leaves $98$ elements to consider. If even a single one is order $9$, it generates a subgroup of order $9$. If any one of them ...


1

Lagrange's theorem tells us if f $G$ is a finite group and $H$ is a subgroup of $G$, then $\vert H \vert$ divides $\vert G \vert$. As a corollary we get that given $a \in G$, $\vert a \vert$ divides $\vert G \vert$ which means that $\vert G \vert = r \vert a \vert$ for some $r \in \mathbb Z$. As a corollary to that we get that $a^{\vert G \vert} = e$ since ...


1

I suppose $p$ is prime. An element in $G = \mathbb{Z}_{p^{2}} \times \mathbb{Z}_{p^{2}}$ is of the form $$g = a^xb^y \ \ \ \ \ x, y \in \lbrace 0,1, \ldots p^2 -1 \rbrace$$ where $a$ is a generator of the first $\mathbb{Z}_{p^{2}}$ and $b$ of the second. We have $ab = ba$ and so the order of $a^xb^y$ is $\text{lcm}(x,y) $. This implies that the number of ...



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