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Hot answers tagged group-theory

8

the structure transport is the composition of two functions, first is hyperbolic tangent, as $$\tanh (x+y) = \frac{\tanh x + \tanh y}{1 + \tanh x \tanh y}.$$ Second is the Mobius transformation $$m(z) = \frac{z+a}{az+1}$$ Note that $$\frac{p+aq}{q+ap} = m\left(\frac{p}{q} \right)$$ Works very nicely. Note that the group identity is $(-a).$ All ...

5

Said differently, proving the claim requires the assertion of the claim. Your reasoning is off here. It isn't accurate to say that proving the claim requires the assertion of the claim. Maybe there is a way to prove it otherwise. (You make a similar mistake in reasoning when you write that "the only way to show that $x=a$ would be...", as Bram28 points out.)...

4

Replacing $n$ with $n-1$ (see the comment by @paw88789), the answer is yes (at least usually). If $n=\prod_{i=1}^kp_i^{e_i}$ for primes $p_i$ then $\langle (1,\ldots,n)\rangle=\langle\sigma_1,\ldots,\sigma_k\rangle$ where $\sigma_i=(1,\ldots,n)^\frac{n}{p_i^{e_i}}$ has order $p_i^{e_i}$. In particular, $S_n=\langle X\rangle$ with $X=\{(1,2),\sigma_1,\... 3 You can't conclude that$1_G\in H$that easily. The only thing we know is that if we have two elements in$H$, then their product is in$H$. We do not know that if the product of two things are in$H$, then each of the factors are in$H$(not even if we already know that one of the factors are in$H$). Instead, to show that$1_G\in H$, you need to consider$...

3

Take $G = \Bbb{Z} \times \Bbb{Z}/2\Bbb{Z}$. Then $\Bbb{Z} \times \{0\}$ is an infinite cyclic normal subgroup of finite index of $G$, but $(0, 1)$ has order $2$ in $G$.

3

Yes, these elements that swap two adjacent strands do generate the braid groups and finite braid groups. In the latter link, such elements are referred to as "elementary braids." Calling them "transpositions" may be confusing, because it's not clear that these correspond to transpositions in the isomorphic subgroup of $S_n$ (maybe they do, but it's not ...

3

$x = a \circ a \circ x^{-1}$ Now, the only way to show that $x = a$ would be if $a \circ x^{-1} = e$ I question this. Yes, it is true that if $a \circ x^{-1} = e$ then we can show $x = a$ ... but it is not at all clear why showing that $a \circ x^{-1} = e$ is the 'only' way to show $x=a$. And yes, you're right, assuming that $a \circ x^{-1} = e$ in ...

2

This is true for example in the following cases. If (a) $G=HZ(G)$, or (b) $H \unlhd G$ and $H \cap G'=1$ (in (b) note that $H \subseteq Z(G)$). This can be generalized by using the concept of isoclinism between groups (write $\sim$), an equivalence relation on the class of groups coarser than isomorphism (see for instance my paper here). In case $G$ is ...

1

Your proof to show that above sequence is short esact is correct, nevertheless define a map $t:K\rightarrow G$, by $t(a)=a$. I have asked similar question (Commutator subgroup of a group of order $8q$, where $q$ is odd prime.) and got the answer, that's why I am giving answer in same pattern: Now to show that $H\subseteq [G,G]$. Let $g=b^k \in H$. Then ...

1

I think Tu's definition is equivalent to the one involving free transitive actions on the fibers. His definition of principal $G$-bundle has two parts: A) we have a fiber bundle $\pi:P\rightarrow M$ with $G$ acting smoothly freely on $P$ AND B) we are told something more about the action: the fiber-preserving local trivializations are $G$-equivariant, ...

1

According to the average order of permutations by Richard Strong, the sum of the orders of all elements of $S_n$ has the following asymptotic: $$n!e^{C\sqrt{\frac{n}{\log(n)}} + O(\frac{\sqrt{n}log(log(n))}{log(n)})}$$ where $C \approx 2.99047$ is a constant.

1

$P(k)$: The identity permutation can be a product of $E_k =2k$ 2-cycles. Math is not magical incantations. Induction is: a) If something is true for a base case. b) if we can prove that if is true for any case it is true for the next one after that, then Conclusion) logically it follows that it is true for all cases after the first one. There is no ...

1

To show the map is well-defined, you have to deal with the possibility of $(K \cap H)a$ being represented in a different way. Specifically, if $a_1$ and $a_2$ belong to the same coset $(K \cap H)a$ in $K$ (i.e. $(K \cap H)a_1 = (K \cap H)a_1$), then they both produce the same coset under this map (i.e. $Ha_1 = Ha_2$). Now, if $(k \cap H)a_1 = (k \cap H)a_1$,...

1

I'll give a slightly different approach that I find more intuitive: (Thanks Derek for pointing out a previous error) We first use the relations to simplify the presentation: $a^b=ac$ and $a^c=a$ implies $a=a^{b^\beta}=ac^\beta$ so $o(c)\mid\beta$. Similarly $(b^{-1})^a=cb^{-1}$ and $b^c=b$ implies $o(c)|\alpha$. Together with $c^\gamma=1$ this gives $o(... 1 The set of endomorphisms for$(\mathbb{R}^3,+)$will form an abelian group under the operation of (pointwise) addition of functions and not with composition, i.e. $$(f+g)(x)=f(x)+g(x).$$ In which case the inverse will be$-f(x)$and commutativity is the outcome of commutative addition in$\mathbb{R}^3$. In general, if we have$(G,+)$an abelian group then ... 1 A transitive$G$-set$X$is defined up to equivalence by the conjugacy class of a point stabiliser$H\le G$(prove by showing that the action of$G$on$X$is equivalent to the natural action of$G$on the cosets of$H$).$S_3$has four conjugacy classes of subgroups, with representatives of orders$1,2,3,6$so up to equivalence$S_3$has four$G$-sets 1 Consider the normalizer of one of the$3$-Sylow groups. You can easily show this is isomorphic to$S_3$. Likewise it is easy to show that the different$3$-Sylow normalizers contain different$3$-Sylow groups. Just as an automorphism permutes the$3$-Sylow groups, it also permutes the normalizers... but if the$3\$-Sylow groups contained within the ...

1

The fallacy is just called a circular argument, but you can definitely not conclude the resulting statement is false. For example, if we restricting to groups of odd order then the statement is true, even though the circular argument is not valid. The problem is that just because you can't find a valid argument doesn't mean one doesn't exist.

1

The problem you are discussing is called 'circular reasoning' or 'assuming what you are trying to prove'. If you are interested in a discussion about mathematical reasoning and possible problems with it more generally, I recommend Chapter 1 of Terry Tao's Compactness and Contradiction. On page 11, he explicitly discusses circular arguments, as well as some ...

1

This is just the composition of inverse change of basis matrices, and so is the identity.

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