# Tag Info

0

As mesel said, this is finding same as maximizing $lcm(a_1,..,a_k)$ for $a_i$ such that $a_1+a_2+...+a_k=7$. This follows from the fact that disjoint cycles commute and the order is then just the number of times a cycle must be composed with itself to make each of these disjoint cycles $e$. But this is a hard question for general $n$- you know that once you ...

0

Your question is equivalent to find nonnegative integers $a_1+a_2...+a_k=7$ such that $lcm(a_1,,,a_k)$ is maximum. To make it maximum,you should choose $a_i$ relativly prime as much as you can.After some try,you can see that answer is $12$.($a_1=3$ and $a_2=4$) To generalize the result for $S_n$ may be challinging.

0

The group $C_2 \times C_2$ is the same as $\mathbb{Z} / 2 \mathbb{Z} \times \mathbb{Z} / 2 \mathbb{Z}$ under addition. The order of $(0,0)$ is of course $1$. The order of all other elements is $2$, because: \begin{align*} (1,1) + (1,1) &= (0,0) \\ (1,0) + (1,0) &= (0,0) \\ (0,1) + (0,1) &= (0,0). \\ \end{align*} In general, the order of an ...

4

Let $G$ be a group. The order of an element $g \in G$ is the smallest $n \in {\mathbb N}, n \geq 1$ such that $g^n = 1$. If a group is cyclic, then there is an element $c$ such that $\{1, c, c^2, \dots, \}$ is the whole group. In the case that this set is finite, you have $c^n = 1$ for some $n \geq 1$ (and after that the sequence repeats); the least such ...

3

The order of an element is the order of the cyclic group generated by the element. I believe it would be more correct to say the period of an element, but to say order i common. In your case $C_2\times C_2=\{e,a,b,ab\}$ with $a^2=b^2=(ab)^2=e$. So the order (more correctly: period) of $a$ is 2 as the cyclic group generated by $s=a$ has two elements. If ...

1

A "square root" of $(13579)$ is $(17395)$. A "square root" of $(268)$ is $(286)$.

2

You have a normal subgroup of order $5$. Your calculations are already sufficient to show that the elements of this group other than the identity don't commute with $\tau$, or indeed any of its powers. So $\tau, \tau^2, \tau^3$ are not in the centre. $1=\tau^0=\tau^4$ is of course in the centre. Suppose we have an element $\rho$ which is in the centre, and ...

1

I found $Z(G)=\{1\}$. Here is my calculation.Let $K=<\sigma>$ and $H=<\tau>$ Since $H$ is cyclic, $H\leq N_G(H)$ and $N_G(H)\cap K=1$ otherwise, $K\leq N_G(H)\implies N_G(H)=G\implies$ $G$ is abelian contradiction.(if $H$ is normal,then $G\cong H\times K$) Thus,we must have $N_G(H)=H\implies Z(G)\leq H$ Since $K$ is cylic $K\leq C_G(K)$ and ...

2

The statement says that $G'$ equals the normal closure N of the set of commutators of the elements of $S$, which is the smallest normal subgroup containing the set of such commutators. To prove this, divide G by N. The quotient group is abelian since it's generators commute. Thus, $N$ contains $G'$. It is also clearly contained in $G'$. Hence, they are ...

0

It appears that the category of finitely cogenerated modules is not closed under extensions in general. Researching this problem led me to this article by Wisbauer. You need to see theorem 6.5 which I believe, when modified for this situation, says that the finitely cogenerated modules of $Mod-R$ are closed under extensions iff $R$ is right ...

0

I find a nice blog post that discusses this problem. I will add the screenshot here just in case the link becomes broken (and make this community-wiki answer).

0

Group is partitioned into conjugacy classes.so any two elements of $G$ you take they belong to two classes and by assumption they commute .so $G$ Is abelian.

3

The stipulations $A\subseteq C$ and $A\cap B=\varnothing$ tell us that $A\subseteq C\setminus B$. The stipulation $\langle A\rangle=C$ tells us that $A$ has to be "big enough" (enough to generate all of $C$). Might as well pick $A=C\setminus B$. To show $\langle C\setminus B\rangle=C$ it suffices to show $B\subseteq\langle C\setminus B\rangle$. Can you show ...

7

The numbers for which there are precisely $1$, $2$ and $3$ groups are classified in the short paper http://www.math.ku.dk/~olsson/manus/three-group-numbers.pdf By Jørn Børling Olsson.

1

Exactly 2 groups. There is a paper, which claims to classify "Orders for which there exist exactly two groups". This link contains the text of the paper in text (!) format. I didn't find a pdf. Disclamer: I didn't check if the proofs in the paper are correct. I also don't know, if the paper was published in any peer-reviewed journal (probably it wasn't). ...

1

Take a look at this wikipedia page on $p$-groups: http://en.wikipedia.org/wiki/P-group The results in the subsection Among groups under the section Prevalence are astonishing. The basic corollary is: Your question is very hard in general.

3

Here are two possibilities for $n$ for which there are precisely two groups of order $n$. $n=p_1p_2 \cdots p_n$ for some $n \ge 2$ and distinct primes $p_i$, such that there is exactly one pair $(i,j)$ with $p_i$ divides $p_j-1$. $n=p_1^2p_2\cdots p_n$ for some $n \ge 1$ and distinct primes $p_i$, where there are no $(i,j)$ with $p_i$ divides $p_j-1$, and ...

5

A partial result: there exists a unique group of order $n$ (i.e. the cyclic group) if and only if $(n,\phi(n)) = 1$, where $\phi$ is the Euler phi function and $(a,b) = \operatorname{gcd}(a,b)$. This is certainly satisfied when $n$ is a prime, but when $n = 15$ we have $\phi(n) = 8$, and since $(15,8) = 1$ there is a unique group of order $15$, even though ...

0

I'm not sure if this is the "nicest" possible description or not, but you might proceed as follows. Case 1. $a\neq 0$ In this case, $a$ is coprime to $p$, so there exist integers $k,l$ such that $ka+lp = 1$. This means that $k(a,b)+l(p,0)=(1,kb)$. This means that we can assume that $a=1$. Note that the subgroup generated by $\{(1,b),(p,0),(0,p)\}$ is ...

2

I have thought of one way of doing this, but there might be easier ways. Let $\mathcal{F}$ be a functor with the properties you describe. Suppose that we have groups $G$, $H$ with homomorphisms $\phi:G \to H$ and $\psi:H \to G$ with $\psi \phi = {\rm Id}_G$. Then $\mathcal{F}(\psi\phi) = \mathcal{F}(\psi) \mathcal{F}(\phi)$ is the identity map on ...

0

Think about the elements that make up $G/N$. It is is the set of cosets of $N$ in $G$. That is, $G/N=\{ Ng \mid g \in G\}$ Is $M/N$ a normal subgroup of this quotient group? It would take a couple of lines to prove it, but you should. Assume that you have proved that $M/N$ is normal in $G/N$. Now what does $(G/N)/(M/N)$ look like? $(G/N)/(M/N)=\{(M/N)g \mid ... 0 Let$K/{\Bbb Q}_p$be a Galois extension and$k$the residue field of$K$. As Galois actions preserve integrality there is an induced action of$G$(the Galois group) on${\frak O}_K$fixing$\Bbb Z_p$; it futhermore fixes powers of the maximal ideal of${\frak O}_K$and so induces an action on$k$over${\Bbb F}_p$, and more generally$G$acts as ... 1 To answer my own question: The four transitive subgroups of$A_7$are as follows: $$C_7, F_{21}, PSL(2,7), A_7$$ 0 Perfect. You don't need to separate cases, observe only that$|xy|=|x|\,|y|$, and then it follows that$\phi=x\mapsto x/|x|$also preserves multiplication. 4 There are two basic ideas: Given$X$, if it is finite then pick any bijection with$\Bbb Z/(n)$, and you have a finite group; otherwise consider$\Bbb Z[X]$, the ring of polynomials whose free variables are elements of the set$X$. We can prove, using the axiom of choice, that$\Bbb Z[X]$has the same cardinality as$X$. Therefore there exists a bijection ... 1 In your particular case, you can get reasonably explicit by transferring the group structure over from a bijection between the irrationals and the reals. It's not hard to write down an explicit enumeration$q_n$of$\mathbb Q$, and then we can map$\mathbb{R}\setminus \mathbb{Q}$to$\mathbb{R}$by$\sqrt{2}/2^{2k+1}\mapsto q_k$and$\sqrt{2}/2^{2k}\mapsto ...

3

You can consider the obvious surjective homomorphism $G_1\times G_2\to (G_1/H_1)\times (G_2/H_2)$. What is its kernel? Then you can use the first isomorphism theorem. This is essentially the same solution as if you just defined a map and checked that it is an bijective homomorphism. But using the 1st isom. theorem you don't need to do this many ...

3

By the earlier answers, all of the elements of order $4$ all have minimal polynomial $x^2+1$, so you can assume that one of them is $A = \left(\begin{array}{cc}0&1\\-1&0\end{array}\right)$. Suppose $B = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ is another matrix of order $4$ in the image. Then, from the structure of $Q_8$, we have ...

1

I suppose that $G$ is acting on a set $\Omega$, to which $\delta$ and $\omega$ belong, $\delta \ne \omega$, and $G_{(\{\delta, \omega\})}$ denotes the stabilizer of $\{\delta, \omega\}$ in the action of $G$ on the subsets of $\Omega$. Then, no, it is not true in general. Consider $G = S_{n}$ acting on $\Omega = \{1, 2, \dots, n \}$. Then ...

1

Hint: define a map $\phi: H \rightarrow HK/K$, by $\phi(h)=hK$ and show that this is a surjective homomorphism with $ker(\phi)=H \cap K$. Then apply the first isomorphism theorem.

2

Suppose that the quaternion group can be embedded into $M_2(\mathbb{R})$. Then it is isomorphic to a finite subgroup of $GL_2(\mathbb{R})$. Since all finite subgroups of $GL_2(\mathbb{R}$ have a faithful real character of degree $2$, this would apply to the quaternion group, too. But this is not the case, see here, a contradiction. Actually, the question has ...

2

There are many solutions of $A^4=I$ in $GL(2,\Bbb R)$, because of conjugation. However it is fairly easy to see that one can nevertheless not embed $Q_8$ in $GL(2,\Bbb R)$. Assuming one has such an embedding, then since $Q_8$ is finite one can find a positive definite bilinear form that is invariant under all elements of $Q_8$ (this a standard averaging ...

4

The inclusion in $(1)$ cannot be an equality for every $i$, in general. If $G$ is nilpotent, then $G = \zeta^i(G)$ for some $i$, and also $[G, G] \subsetneq G$, so $[\zeta^{i+1}(G), G] \subseteq [G, G] \subsetneq \zeta^i(G)$. A nice way to see $(2)$ has already been addressed in the comments above.

2

(1): Consider $G = D_8 \times C_2$, where $D_8$ is dihedral of order $8$ and $C_2$ is cyclic of order $2$. For this group, $\zeta^2 = G$ and $[G,G]$ is a proper subgroup of $\zeta^1$. (2): If $N$ is characteristic in $G$, then we have the map $\operatorname{Aut}(G) \rightarrow \operatorname{Aut}(G/N)$ defined by $\phi \mapsto \hat{\phi}$, where ...

4

Usually the way to approach matrix polynomial equations (of one variable) is to figure out what the minimal and/or characteristic polynomial of the solution matrices would be. Note that any conjugate of a solution to $p(A)=0$ is also a solution, so if there is one solution there will be infinitely many (except in cases where the only solutions are scalars). ...

3

More interesting is the converse, the following is true: if $G/Z(G)$ is cyclic then $G$ is abelian.

6

The center of $G$ consists of those elements $g \in G$ such that $gh = hg$ for all $h \in H$. In an abelian group, this relation always holds, so $Z(G) = G$. It follows that $G / Z(G)$ is the trivial group, consisting of only a single element (namely, the coset $Z(G)$). Every group with one element is cyclic.

1

The idea: $~A/B\cong C\iff$ there is a surjection $A\to C$ with kernel $B$. For $H/(H\cap K)\cong HK/K$, what groups need to be $A$, $B$ and $C$?

0

You are given that $G$ is $4$-transitive on $\Omega$, and the triples $(a,b,\omega)$ and $(c,d,\omega)$ each consist of three distinct elements (given, as stated, that $a\neq b$ and $c\neq d$). Therefore, there is an element $g$ in $G$ for which $(a,b,\omega)^{g} = (c,d,\omega)$. In particular, $\omega^{g} = \omega$, so that $g$ belongs to $G_{\omega}$.

3

Note that $|0|=0$, so certainly $\{\alpha\in \Bbb C:|\alpha|=1\}=\{\alpha\in\Bbb C^\times:|\alpha|=1\}$

2

I will restrict myself only to Lie groups, since the world of groups outside of this class is way too large and, I do not think, there is a good answer in this context. I will also restrict to simply-connected Lie groups so that the answer is reasonably neat. (This is not a very good reason, but this will keep my answer reasonably brief, I did not think ...

1

This is essentially showing that if a group is of order 4 and not cyclic, it is the Klein 4 group isomorphic to $C_2 \times C_2$ - all groups of order 4 are isomorphic to one of these groups. So to prove this without Lagrange's Theorem, we can suppose $o(x)=3$, else $o(x)=1$ if $o(x) \neq$, so x=e and this is trivial. So if $o(x)=3$, the group can be ...

2

First of all, by definition the order of an element $x\in G$ is the smallest positive number $n$ such that $x^n=e$. So we might as well rule out $O(x)=0$. ($O(x)=0$ doesn't even make sense anyway...what is $x^0$ supposed to mean?) If $O(x)=1$, this by definition means that $x^1 = x = e$. In this case $x^2=e$ is trivial. If $O(x)=2$, then $x^2=e$; this is ...

2

Theorem Let $G$ be a non-trivial finite group. Then the following are equivalent. (a) For each pair of subgroups $H_1$ and $H_2$ of $G$, $H_1 \cup H_2$ is a subgroup (b) $G$ is cyclic of prime-power order. Proof (b) $\Rightarrow$ (a) follows from the fact that in a cyclic group there is a unique subgroup of order $d$ for each divisor $d$ of $|G|$. (If $G$ ...

1

So, hints: What can you say about $xyx^{-1}$? By the isomorphism theorem $G/\ker f \cong \operatorname{img} f$. a. If $f$ is one-to-one, $\ker f = 1$ and so … b. If $f$ is onto, $\operatorname{img} f = H$ and so … c. Use a. and b. If you don’t know the isomorphism theorem, you can take arbitrary elements $x, y$ in $G$ or $H$ and have a look at images ...

1

Let $G=S_3=\{(1),(1,2),(1,3),(2,3),(1,2,3),(1,3,2) \}$ and $$H_1=\{(1),(1,3) \}$$ $$H_2=\{(1),(1,2) \}$$ $$H_1\cap H_2=\{(1) \}$$ $$H_1\cup H_2=\{(1),(1,2),(1,3)\}$$ The union is not group since it is not closed as $(1,2)(1,3)=(1,3,2)\notin H_1\cup H_2$ for proof of inersection is a subgroup,just write; if you can take an element from a set you can do ...

0

You can see the first from the definition of subgroup or the subgroup criterion quite easily. It's quite easy to make $H_1\cup H_2$ not a group... Consider two subgroups that intersect trivially. What happens when you add non-identity elements from these two subgroups?

3

Use the definition of a subgroup. You know that they both share the identity element, and for any element they share, they must also share the inverse. Can you take it from here?

1

If $x,y\in S_n$ are both odd permutations the they are both the product of an odd number of transpositions, say $m,n$ respectively Then $xy$ is the product of $m+n$ transposition which is even hence result.

1

Nick's answer is ideal. Here's another, less ideal, solution. If $G$ is infinite and cyclic, say $G=\langle x \rangle$ then $\langle x^2 \rangle\subsetneq G$. Otherwise there is some element for which $\langle x \rangle\subsetneq G$. So $G$ is finite. If $G$ has order $p^r m$ where $p\not|\ m$ then it has a subgroup of order $p^r$ by Sylow's Theorems. So ...

Top 50 recent answers are included