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6

A $p$-group is a group where the order of every element is a power of $p$. A direct consequence of this is that if the order of every element is a power of $p$, then the order of the group must be a power of $p$. This follows from Lagrange's theorem: the order of a subgroup of a group divides the order of the group. So, one can say that a $p$-group is a ...


1

When $n=2$, the braid group $B_n$ is just the free group on one generator $\sigma$, and the representation is determined by the matrix via which $\sigma$ acts on $\mathbf{C}^2$; for $z=-1$ this is $$\sigma \mapsto \left( \begin{matrix} 2 & -1 \\ 1 & 0 \end{matrix} \right). $$ This matrix is conjugate to $$\left( \begin{matrix} 1 & 1 \\ 0 & 1 ...


1

Hint: Take the characteristic homomorphism $\rho: \mathbb Z \to F$, where $\rho (n) = n 1$. Notice that $F$ is a domain and its characteristic must be a prime $p$ and $\rho (\mathbb Z) \simeq \mathbb Z_p$, because its kernel is $I(p)$, that is, the ideal generated by $p$ is the smallest subfield of $F$.


0

Yes, the set of values $0,1,1+1,1+1+1,\dots,$ is closed under multiplication and addition (why?), and it must be a subgroup under addition. Since it is necessarily cyclic and finite, it is of the form $\mathbb Z/n\mathbb Z$. Use field properties to show that $n$ must be prime.


0

Yes it's true that every finite field contains $\mathbb F_p$ for one value of $p$. In fact every field contains either $\mathbb F_p$ for some $p$, or $\mathbb Q$.


1

HINT: Given $f\in F$, consider the following functions: for each $m\in\Bbb N$ let $g_m\in F$ be defined by $$g_m(k)=\begin{cases} f(k),&\text{if }k\ne m\\ f(k)+1,&\text{if }k=m\;. \end{cases}$$


1

Let $n=|\sigma|$, then $n|p$ so $n=1$ or $p$. If $\sigma\ne e$, then $|\sigma|=p$. Now since $1\le r<p$, $r$ and $p$ are coprime and so $|\sigma^r|=p$. Finally let $i\in\{1,\ldots, p\}$. The orbit of $i$ consists of $i,\sigma(i),\sigma^2(i),\ldots,\sigma^{p-1}(i)$, therefore $\sigma$ is a $p$-cycle.


2

For the first question, it is possible for $G$ to be finitely presentable. The free group $F$ on two generators $a,b$ is the semidirect product of the normal closure of $a$ in $F$ with the infinite cyclic subgroup generated by $b$. (This is the same example that you gave. In your example, $F$ is freely generated by $t$ and $x_0$.) The answer to the second ...


3

Since there is an answer up, I would do $$ax^2a=x^3$$ $$x^2=a^2x^2a^2=ax^3a=x^3axa$$ $$axa=x^{-1}$$ $$ax^2a=x^{-2}=x^3$$ Leaving you with some bits to fill in.


0

Exercise: let $G$ be a simple group and fix a field $K$. Then every nontrivial linear representation of minimal dimension is faithful and irreducible unless $G$ is cyclic of order $p$ and $p=0$ in $K$. In particular, if the simple group $G$ admits a nontrivial linear representation over $K$ (e.g., $G$ is finite), then it admits a faithful irreducible ...


5

$a^{-1}x^2a=x^3 \implies a^{-1}x^4a=x^6 \implies a^{-1}x^6a=x^9$ but $x^6=x^3x^3=(a^{-1}x^2a)(a^{-1}x^2a)=a^{-1}x^4a \implies a^{-1}(a^{-1}x^4a)a=x^9 \implies a^{-1}a^{-1}x^4aa=x^9$ now using $a^2=a^{-2}=e$ we have $x^4=x^9$ so $x^5=e$ and because $5$ is prime $x=e$ or $x$ have order $5$.


1

(We only consider the complex representation) Suppose that $G$ is non-Abelian. Then there is an irred repr. $\rho$ s.t. it is $d$-dim'l, $d>1$. By dimension theorem, we get $d=p,q$, or $pq$, but we know that $d$ must be $p$; otherwise by the property that $\sum_{\chi'\in\text{Irr}(G)}d_\chi^2=|G|$, we will get $d^2>pq$ because of the condition ...


2

Every finite group is isomorphic to a subgroup of $S_n$. Now take $G$ a subgroup of odd order included in $S_n$. Let $g\in G\subseteq S_n$, write its unique decomposition into cycles : $$g=c_1...c_r\text{ then } \epsilon(g)=\epsilon(c_1)...\epsilon(c_r) $$ Now you know how to compute the signature of a cycle, namely if $c$ is a cycle of length $l$ then ...


2

Hint Say, left multiplication by any element $g$ of a finite group $G$ permutes the elements of $G$ (and by uniqueness of the identity element this permutation is distinct for each element $g$), so we can always regard $G$ as a subgroup of $S_G \cong S_{\# G}$. For any $m$ there is an injective homomorphism $S_m \hookrightarrow A_{m + 2}$. Remark Putting ...


1

As noted, the quotient is an abelian group of order $48$. The element $x:=(1,01)$ is of order $24$ in the quotient since $$ 24(1,0,1)=(24,0,24)\equiv(24,16,32)=8(3,2,4) $$ but if the order of $x$ was less then $24$ then by Lagrange it would have an order dividing $24$. Since $24=2^{3}\cdot3$ $|x|\mid24$ and $x\neq24$ imply $|x|\mid12$ or $|x|\mid8$ but ...


2

Yes. It follows easily from the fact that maximal subgroups of finite supersolvable groups have prime index.


1

An answer in the negative for this question would immediately prove the Smith conjecture: Does the finite part of the automorphism tower of a group G eventually repeat? (Smith 1964) I have spent some time wrestling with this conjecture, and the pattern I observed was that this did occur and typically very rapidly (only a few steps in, I was using GAP to do ...


1

(a) If $H$ is a proper subgroup of $G$, then the order of $H$ is a divisor of $2n$. If $n$ is prime, the only possible divisors of $2n$ are $1, 2, n$ or $2n$. Since the last one is not possible,.... (b) Let $x \in G$ be $x \neq e$. Then the order of $x$ is a divisor of $n^2$, thus $1,n$ or $n^2$. The order cannot be $1$. If the order of $x$ is $n$ then ...


1

First you need to check whether it is a group. Just checking closure under addition is not enough. First closed under addition (done), second associativity ($(ab)c=a(bc)$), then existence of identity ($(1,1)$), then existence of inverse ($(a^{-1},b^{-1})$). Communitivity is OK, but it seems not to be the natural way people write. $(a,b) \oplus (c,d) = (c, ...


0

You forgot to find a inverses for $(a,b)$ and you forgot to find an identity, you also forgot to prove the operation is associative. Also not this method for creating a new group from two old ones can be generalized and is called the direct product (sometimes called direct sum when working with abelian groups).


1

To show that $f$ is a homomorphism, we must show that $f((a,b) \oplus (c,d))=f((a,b))+f((c,d))$. So, \begin{align} f((a,b) \oplus (c,d)) &= f((a+c,b+d))\\ &=a+c-(b+d)\\ &=a+c-b-d\\ &=a-b+c-d\\ &=f((a,b))+f((c,d)) \end{align} So, $f$ is a homomorphism. To show $f$ is onto (surjective), we must show that for every $y \in \mathbb Z$, there ...


1

|ab| does not in general divide lcm(|a|,|b|). See the earlier post, with a link to a nice proof. Order of product of two elements in a group


0

This is the direct product of four copies of $C_2$ (the cyclic group with $2$ elements). So you can think of this group as $\mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z \times \mathbb Z/2\mathbb Z$. The elements are $(a,b,c,d)$ with $a,b,c,d \in \{0,1\}$


2

Unfortunately, no. In fact, it's even worse! The quaternion group $Q_8$ has subgroups of all possible orders, and every subgroup is normal, but the group isn't abelian! More generally $p$-groups, groups whose orders are a power of a prime, will have subgroups of all orders, yet need not be abelian. There is a partial converse to Lagrange's theorem, about ...


10

The answer to your question is no, that is not possible. This is a consequence of the following slightly more general result. Let $d(n)$ is the number of divisors of $n$. Theorem Let $G$ be a finite group of order $n$. Then for any divisor $m$ of $n$, there exist at least $d(m)$ subgroups of $G$ of order dividing $m$. Proof We prove the statement by ...


1

In short, you are correct. $(15)(36)$ is the right representation of $a$ as a product of disjoint cycles, and $a^{24}$ is the identity while $a^{25}$ is just $a$ again, because $a$ has order $2$. Note that it doesn't matter if we write $(51)(63)$ or $(15)(36)$ - in general cyclic permutations represent the same cycle.


2

I like the construction Tobias has posted. Here is a more hands-on method. The Sylow theorems tell us that a group of order $21$ has a normal subgroup of order $7$. Let this be generated by $a$, and let $b$ be an element of order $3$. Then, because $a$ generates a normal subgroup, we have $bab^{-1}=a^r$ for some $1\le r \le 6$. If $r=1$ the group is ...


3

Here is a way to construct the nonabelian group of order $21$ without referring to semidirect products. Let $F$ be the field with $7$ elements, which we will identify with the set $\{0,1,2,3,4,5,6\}$. Note that the elements $H = \{1,2,4\}$ form a subgroup of the units of $F$ of order $3$. Now define $G = \{f: F\to F\mid f(x) = ax + b,a\in H, b\in F\}$. ...


0

If I understand the question you are being asked, your scenario is the following: (1) The set $S$ is your domain; in particularly, $S = \mathbb{R^+}$. (2) The binary-operation you are working with is $a \circ b = 4ab^2$ (3) You are being asked to find the identity for this operation. The difficulty is here is that you will not find an identity ...


0

For finite groups, your guess is correct; not sure about the infinite case. For a proof, let $G = HK$, suppose that $H \unlhd G$, $K \unlhd G$, and that $G/H \cong K$. To show that $G\cong H \times K$, we need only show that $H \cap K =1$. We appeal to the diamond isomorphism theorem, which says in this case that $G/H=HK/H \cong K/H \cap K$. But since $G/H ...


0

Notice that $D_4/Z(D_4)$ is a group of order 4 (in particular, its the Klein 4-group), and thus abelian. The map in question is simply conjugation in an abelian group, so it is the identity.


0

In general, it is not valid. $a^2=e$ does not imply $a=e$. Even in $D_4$ it is not true I'm afraid. The condition and conclusion is right here though. $\overline{\phi}=\mathrm{Id}$ just means that the quotient group is commutative. Also note that $$D_{2n}=\langle r,s |r^n=s^2=e, s^{-1}rs=r^{-1}\rangle$$ or in your notation it means ...


2

We need these conditions: $1.)$ $G=HK$ $2.)$ $H$ and $K$ normal in $G$. $3.)$ $H \cap K= \{1\}$ Then $G=HK \cong H \times K$


1

Your second guess is a correct description of $\Bbb Q(\pi)$ (as long as you specify that the denominator is not the zero polynomial). The first guess describes the ring $\Bbb Q[\pi]$, which does not have inverses. In particular, $\pi$ has no inverse since $\pi f(\pi) \ne 0$ for any polynomial $f\in \Bbb Q[x]$. It's not hard to see that these rational ...


0

It is your second option. The first is called $\Bbb Q[\pi]$. In the example they make $F=\Bbb Q$ and $\alpha=\pi$.


9

If $\pi\in S_5$ is a permutaion of $\{1,2,3\}$, then either $\pi$ or $\pi\circ (4\,5)$ is an even permutation, i.e., $\in A_5$. This allows us to embed $S_3\to A_5$. Or have fun with geometry: $A_5$ is the symmetry group of the dodekahedron. It is possible to select $4$ of its $20$ vertices that make up a regular tetrahedron $T_1$. A rotation of the ...


2

Every element in a group has an inverse. In your case, there must be some element $x \in \{a,b\}$ for which $b * x = a$. To see this another way, note that no two elements in a row are equal. This is because if $b * x = b * y$, then by multiplying on the left by $b^{-1}$, we have $x = y$.


1

Although you said that you didn't want a yes/no answer, my hint to you is that the answer is yes, so you should just try to construct a subgroup, $H$, of order 6. Look for elements $\sigma$ and $\rho$ of order 2 or 3. A priori there are two possibilities for $H$ - namely $\mathbb Z_6=\langle\rho, \sigma | \sigma \rho \sigma ^{-1}=\rho, \quad ...


2

This is kind of a silly example: $\mathbb{Z} \subseteq \mathbb{Q}$, but the Krull dimension of $\mathbb{Z}$ is one while the Krull dimension of $\mathbb{Q}$ is 0.


1

Look at the answer to this StackExchange question: http://math.stackexchange.com/questions/132729/a-free-submodule-of-a-free-module-having-greater-rank-the-submodule (and also the MO discussion referenced therein).


0

If $G$ is finite, then it is uniquely $p$-divisible if and only if $p$ does not divide $|G|$. (This is a generalisation of your example.) See for example: http://groupprops.subwiki.org/wiki/Kth_power_map_is_bijective_iff_k_is_relatively_prime_to_the_order


6

$2014 = 2 * 19 * 53$ As $2, 19, 53$ are all distinct, there is only one Abelian group of order $2014$ up to isomorphism, which is $\mathbb{Z}_{2014}\simeq\mathbb{Z}_{2}\oplus\mathbb{Z}_{19}\oplus\mathbb{Z}_{53}$. Also you can proceed this way without using the classification theorem. By Cauchy's theorem there are elements of $2, 19, 53$. Suppose they are ...


0

So let $S_i$ be the words generated beginning with $A$ or $A^{-1}$ and let $T_j$ be the words that beginning with $B$ or $B^{-1}$, because $S_i$ and $T_j$ can have two options, both $m$ and $n$ are both $2$?


1

The group $SL_2(\mathbb{R})$ has a natural but nonfaithful action on the projective space $\mathbb{R}P^1$. Letting $M \in SL_2(\mathbb{R})$ and letting $\ell_{\vec v} \in \mathbb{R}P^1$ be the line through the origin with a direction vector $\vec v$, the action is given by $$M \cdot \ell_{\vec v} = \ell_{M \vec v} $$ The kernel of this action is all ...


4

Hint: By Langrange's Theorem if $H < G$ then $|H|$ divides $p$. As $p$ is prime then we must have $|H| = 1$ or $|H| = p$. What can you conclude from that?


1

A group of order $p$ has $2$ subgroups: the trivial subgroup and the entire group.


0

Let $H$ be elementary abelian of order $p^n$ for an odd prime $p$, and let $G = H \rtimes \langle t \rangle$, with $t^2=1$, where $t$ inverts every element of $H$. (This is soemtimes called a generalized dihedral group.) Then generating sets for $G$ have size at least $n+1$, but $|G/[G,G]|=2$.


3

The use of Tietze transformations to prove that $\langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2$ is correct. There are no mistakes in that. The mistake is to claim that $\langle a,b;(ab)^2\rangle$ is a presentation of the projective plane. In fact, the relation $\langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2$ together with the Classification ...


0

For a concrete example, groups with a non-trivial center acting on their set of subgroups via conjugation work well (in fact, I think Hagen von Eitzen gave you a description of all non-faithful group actions). In the dihedral groups acting on a regular $2n$-gon with $n$ even, the half-rotation is in the center of the group, and fixes every subgroup.



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