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3

The group theoretic interpretation is as the numbers game for the Coxeter group of type $\widetilde{A}_4$, with generators $s_1,s_2,s_3,s_4,s_5$, where the relations are $$s_1^2=s_2^2=s_3^2=s_4^2=s_5^2=1\mathrm{,}$$ $$(s_is_j)^2=1$$ if $i\not\equiv j\pm 1\pmod{5}$, and $$(s_is_{i+1})^3=1$$ for all $i$, where $i+1$ is taken modulo $5$. This is an infinite ...


3

There is a version of the tensor product for nonabelian groups, but this notion is much more specialized. See http://www-irma.u-strasbg.fr/~loday/PAPERS/87BrownLoday%28vanKampen%29.pdf, section 2. In the construction at some point you do a mod out, which you cannot do in general if you do take the free group instead of the free abelian group. (You see a free ...


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 $S_{30}$ ...


1

This proof works. As @bof noted in comments, there is a more direct approach, not using the more advanced result about the normal subgroups of $S_n$ for $n>4$. @hof's answer works for all $n$.


1

If $G$ is a group of order 30, then the number of Sylow 5-subgroups is congruent to 1 modulo 5 and divides 30, so it must be 1 or 6. And the number of Sylow 5-subgroups is congruent to 1 modulo 3 and divides 30, so it must be 1 or 10. If there is a unique subgroup of order 3 or 5, it is normal and we are done. If not, there are 6 subgroups of order 5 and 10 ...


1

By the Sylow theorems, the number of Sylow $p$-subgroups of the group is congruent to $1$ mod $p$ and divides the order of the group. Suppose $n=kp+1$ and $n\mid q$. Then since $q$ is prime we in fact have $kp+1=q$ or $k=0$. Assume $k\neq 0$. Then $q-1=kp$, meaning that $p\mid q-1$. Since we assume that this is not the case, we must have that $k=0$, hence ...


1

Let $G$ a finite $p$-group. Define $\mathscr{P}_1(G) = G$, and $\mathscr{P}_{i+1}(G) = [\mathscr{P}_i(G), G]\mathscr{P}_i(G)^p$ for $i > 1$. Then there exists an integer $c$ such that $\mathscr{P}_{c+1}(G) = 1$. If $c$ is the smallest such integer, we say that $G$ has exponent-$p$ class $c$. Note that if $G$ has exponent-$p$ class $c$, then $G$ has ...


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 ...


0

Let's $A=\{e,a,a^2\}$ and $B=\{e,b,b^2\}$ are cyclic subgroups of order $3$ such that $A\cap B={e}$, then $H=AB$ has order $9$, otherwise we would have $a^ib^j=a^mb^n$ for some $i,j,m,n$, which is contradicts that $A\cap B={e}$. Now we only left to prove that there exist such $A,B$. But it follows from the fact that there can be only one subgroup of order ...


1

De Polignac's formula is a result in combinatorics, so I would think you're able to use it without proof on an algebra problem, because you don't have the tools to prove it right now. It's not as complicated as it looks. It just lets you count the multiplicity of a prime in $n!$, which is the order of $S_n$. You can do this without any fancy formulas. In ...


1

A set of elements $g_1, \ldots, g_n$ are called generators for a subgroup $S$ if $S$ is the smallest subgroup that contains $g_1, \ldots, g_n$. This is equivalent to saying that $S$ is the intersection of all the subgroups that contain $g_1, \ldots, g_n$. It's also equivalent to saying that $S$ is the subgroup consisting of all elements which can be ...


3

Note that any reflection has order $2$, so any group of symmetries which includes a reflection must have even order. You can generalise your solution for $5$ to any odd order.


8

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


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 ...


0

Note that $99=3^2\cdot 11$ so any $3$-Sylow has order 9. By the Sylow theorems, the number of $3$-Sylow subgroups, $n_3$, satisfy $$ n_3\equiv 1\mod 3\qquad\mbox{and}\qquad n_3|11$$ Thus, there is one $3$-Sylow subgroup (and it necessarily has order 9).


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 + ...


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 ...


-1

If it's abelian then you know it's a product of cyclic groups. You know that the order of these cyclic groups can be $3, 9, 11, 33$, or $99$ and their product must be $99$. You've already seen that if $99$ is in their that you're done, and if $9$ is in their then you're done. So your left with a product of cyclic subgroups of orders $3, 11$, or $33$ and ...


1

$|G|=|Ker(\varphi)|\cdot|Im(\varphi)|$ and $|Im(\varphi)|\cdot |H:Im(\varphi)|=|H|$. Hence their common divisor is at least $|Im(\varphi)|$.


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

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

Actually, you have an error in understanding the category $\mathcal{G}$ obtained from $G$. It is not the full subcategory of $\mathbf{Grp}$ whose only object is $G$: i.e. it is not the category with one object whose arrows are endomorphisms of $G$. Instead, it is the abstract category with one object whose arrows are elements of $G$. The product of arrows ...


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 $\mathbb Z \subset 2\mathbb Z$.


0

From Blah's answer. Let $\sigma\,:G'\to G'/N$ be the canonical homomorphism, and $\psi=\sigma\circ\phi$. Since, $|G'/N|=2$ and $|G|>2$, cannot be injective. So $\ker\psi=G$. In other words, $\psi\left(G\right)=\{eN\}=N$. This implies $\sigma\left(\phi\left(G\right)\right)=N$, which means $\phi\left(G\right)\subseteq N$.


1

Since $A\leq B$, as a set $A\subseteq B$. Also, $[G,A]=[G,B]$ yields $$\frac{|G|}{|A|}=\frac{|G|}{|B|}\iff|B|=|A|.$$ So $A$ is contained in $B$, but they have the same number of elements. Therefore, $A=B$.


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

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

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 ...


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

For 3, in a group the relation $\neg$ is called conjugation. If $x\neg y$ for some $x,y\in S$, then there exists $s\in S$ such that $s*x=y*s$ and so $s^{-1}*y=x*s^{-1}$, which implies that $y\neg x$.


0

${\mathbb Z}_n$ is a monoid (a semigroup with 1) under multiplication. 0 is its absorbing element. Absorbing element can’t be invertible unless the monoid is trivial (1 = 0); that’s why for $n>1$ there is no group.


0

Hint : From Cayley's theorem, we know that $G$ can be embedded inside $S_{|G|}$. So to embed $G$ inside $GL_n(\Bbb F_p)$, you have to just find an injective homomorphism from $S_{|G|}$ to $GL_n(\Bbb F_p)$. (note that this is not possible for all $n, p$)


3

Your claim is NOT true. $M = \mathbb{Z}, N = \mathbb{2Z}, M' = \mathbb{8Z}, N' = \mathbb{4Z}.$ All are considered modules over $\mathbb{Z}.$ Then $\frac{M}{M'} \oplus \frac{N}{N'} \cong \mathbb{Z}_8 \oplus \mathbb{Z}_2.$ On the other hand $\frac{M}{N'} \oplus \frac{N}{M'} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_4.$ So this two are not isomorphic. The mistake ...


3

Let $aa'=e$; consider $a''$ such that $a'a''=e$. Then $$ a=ae=a(a'a'')=(aa')a''=ea''=a'' $$ and therefore $a'$ is both a left and right inverse of $a$.


2

In general, an action of a group $G$ on a set $X$ is a group homomorphism from $G$ to the group $S_X$ of permutations of the set $X$. This means that we send each group element $g \in G$ to some permutation of the elements in $X$, so each group element "acts" on $X$ by permuting its elements in some way. Usually, we write $g \cdot x$ to denote the element of ...


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 ...


1

i) Take $h^2 \in G_2$ and $ g \in G $, then $$g^{-1}h^2 g = (g^{-1}hg) (g^{-1}hg) \in G_2$$ This means that $ G_2 \triangleleft G$ ii) Take $\overline{g} \in G/G_2 $, thus $$\overline{g}^2 = \overline{g^2} = 1_{G/G_2} $$ i.e all elements in $G/G_2$ have order $ \leq 2$. iii) Take $h^2 \in G_2 $, thus $\phi(h^2) \in \text{img}(\phi) $ but $$ ...


1

(i) Let $g' \in G_2.$ Then $g' = g^2$ for some $g \in G.$ Now for every $h \in G$, $hg'h^{-1} = (hgh^{-1})^2 \in G_2.$ (ii) Let $gG \in G/G_2.$ Then $(gG)^2 = g^2G.$ (iii) Let $\phi : G \rightarrow H$ has the above stated property. Then for each $x \in G, \phi(x)^2 = e_H.$ So in particular, for $x \in G_2, \phi(x) = e_H.$ Hence, $G_2 \subseteq $ Ker$\phi.$ ...


1

I will show the one direction and the other is an exercises for you. Note that for a normal group $N$, $gN=Ng$ thus for $gn$ there is a uniqe $n_2$ such that $gn=n_2g$. Let $H$ be a normal subgroup of $G$ and $K=H\times H$ $|K:Stab(x)|$ is the size of orbit containing $x$. $$hxh'^{-1}=x$$ $$xh_2h'^{-1}=x$$ $$h_2=h'$$ Note that for any $h$, there is a ...


0

Hint. $hxh'^{-1}=h(xh'^{-1}x^{-1})x$.


2

It seems OP's question (v1) is essentially asking: Given a Lie group $G$, with a Lie group representation $R:G\to GL(V,\mathbb{F})$, with corresponding Lie algebra $L={\rm Lie}(G)$, and with a corresponding Lie algebra representation $\rho :L\to {\rm End}(V,\mathbb{F})$, would $$\tag{1} \forall g\in G, x\in L:~~\rho({\rm ...


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


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

http://math.stanford.edu/~dford/math109/solution1.pdf Adjust the 1 in the link for more. And you can also google for some answers, there are more people who have posted their questions from the book (formulated in the same way as in the book).


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 ...


0

Hint: $$|X|=\sum_{x\in X}'\left|\mathcal Orb(x)\right|=|A|+\sum_{x\in X\setminus A}'\left|\mathcal Orb(x)\right|$$ Where $\;\sum'\;$ means sum over disjoint orbits. Now just remember that $$\left|\mathcal Orb(x)\right|=[G: G_x]\;,\;\;G_x:=\{g\in G\;:\;gx=x\} =\text{the isotropy or stabilizer group of}\;x$$


1

Suppose $G$ is nonabelian. Conjugation induces a map from $G \to \operatorname{Aut}(G)$ that descends to a map $f:G/Z(G) \to \operatorname{Aut}(G/Z(G))$. Since finite $p$-groups have nontrivial center, $Z(G) = \mathbb{Z}_p$. Hence $f:\mathbb{Z}_p \to \mathbb{Z}_p^\times$, and so must be trivial. The result follows.


1

The center $Z(G)$ is non trivial. If $|Z(G)|=p^2$ then it is obvious. If $|Z(G)| = p$ then $|G/Z(G)| = p$ and so $G/Z(G)$ is cyclic because every group of order $p$ with $p$ prime is cyclic. This implies that $G$ is abelian. Infact Suppose $G/Z(G) = \langle{\overline{g}}\rangle$ and $a, b \in G$. Thus in $G/Z(G)$ we have$$\overline{a} =\overline{g}^h $$ ...


9

Using a lot less machinery: Such a representation must land in $SL_2(\mathbb C)$, and finite subgroups of $SL_2(\mathbb C)$ are completely classified (they are either cyclic, dihedral, tetrahedral, octahedral or isocahedral). None of these groups are simple, unless cyclic of prime order. One can prove this classification by letting a finite subgroup $G$ of ...



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