# Tag Info

3

Apart from the trivial group, no. If $x \in \mathbb{R}\neq 0$ then $x, x + x ,x + x + x ,\ldots$ is an infinite sequence of distinct elements of any group containing $x$. Hence any subgroup of $\mathbb{R}$ containing a non-zero element is infinite.

3

Certainly the Theorem of Cauchy would do it - if $q$ is a prime other than $p$, dividing the order of $G$, then there would be an element of order $q$, a contradiction.

2

The comutativity is equivalent to $$y_1 y_2= y_2 y_1 \forall y_i\in K$$ $$x_1 \phi ( y_1 ) ( x_2 ) = x_2\phi ( y_2 ) ( x_1 ) \forall x_i \in H y_i \in K$$ The $K$ must be abelian, from the second equation we take $$\phi (y_2 ) ( x_1 ) = x_2^{-1} x_1 \phi ( y_1 ) ( x_2 )$$ So $\phi (y_2 ) ( x_1 )$ does not depend on the value of $y_2$, hence $\phi$ is a ...

2

Consider $\{id, (12)(34)\},$ $\{id, (13)(24)\}$, and $\{id, (14)(23)\}$. There is no cyclic subgroup of order $6$ in $S_4$, hence no subgroup of order $6$ in $S_4$ isomormorphic to $\mathbb Z_6$.

2

All groups here will be abelian, with additive notation. Definition. A group $G$ is said to be torsionfree if, for all $x\in G$, $x\ne0$ and $n>0$, we have $nx\ne0$. A group $G$ is said to be torsion if, for all $x\in G$, there exists $n>0$ such that $nx=0$.  A group can be neither torsion nor torsionfree. An example is ...

2

I guess you are mixing $\mathbb{Z_8}$ and $U(8)$. $1$ is identity in $U(8)$ as operation is multiplication here. But in $\mathbb{Z_8}$ operation is addition, so $0$ is the identity and $1$ is the generator in $\mathbb{Z_8}$ as you can have any elements less than $8$ by adding $1$. No $U(8)$ is not cyclic. To see it pick an element of $U(8)$ and keep ...

2

This is incorrect, as Alastair's counterexample shows. Two related statements that are correct are: If $\sigma$ is an $n$-cycle (more generally, an $m$-cycle for $m | n$) then $\sigma^n = \iota$. For any finite group $G$, any element $g \in G$ satisfies $g^{\# G} = \iota$, and so for $\sigma \in S_n$, $\sigma^{n!} = 1$.

2

This is an extended comment Qiaochu answer; so extended indeed that it is larger than its progenitor. (Warning: I largely made it up on the go, so there may be mistakes.) First, the concrete. Consider the "mixed Heisenberg group": namely, the set of "matrices" $\begin{pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix}$ ...

1

Recall that a group $G$ is said to act on a set $X$ if each $g \in G$ can be thought of as a bijection $X \to X$ (and we need that the identity $e \in G$ is the trivial bijection and that these maps satisfy associativity $(gh)(x) = g(h(x))$ for $x \in X$). Any bijection of a set can be thought of as a rearrangement or a permutation of the set $X$, so what ...

1

'*' indicates dual or weighted lattices. See this pdf. But, to fulfill this answer, I may assure you that $A_n\;^*$ means the lattice of vectors having integral inner products with all vectors in $A_n:A_n\;^*=\{x\in R^n |(x,r)\in Z \;for\;all\;r\in A_n\}$. And, that,$A_n\;^*\;^*=A_n$.

1

Let he order of the group be $n$ since the group is cyclic we can assume it is of the form $g^k$, we then need to find the smallest value $l$ so that $(g^k)^l=1$ notice this is $g^{kl}=1$. This only happens if $n$ divides $kl$ by how the cyclic subgroup works. Therefore we need to find the smallest postive $l$ so that $n|kl$ this is clearly going to be a ...

1

Recall that $S_2$ is a cyclic group of order 2. A subgroup of $S_4$ has order 2 if and only if it is generated by a permutation of order 2. A permutation of order 2 in $S_4$ is either of type $(i,j)$ (a transposition) or of type $(i,j)(k,\ell)$ (a product of two disjoint transpositions). The number of such generators is exactly 9: the six transpositions ...

1

If $H$ normalizes $P$, then $HP$ is a subgroup and $|HP|=|H||P|/|H \cap P|$, whence $HP$ is a $p$-group. Since $P \subseteq HP$, $|P| \leq |HP|$.

1

if $H$ and $P$ are Sylow subgroups then $H \cap P$ also has order a power of $p$. since the order of $H$ is also a power of $P$ then $(HP:P)=(H:H\cap P)$ is a power of $p$.

1

If $\phi:K\to{\rm Aut}(H)$ is not trivial then $\phi_k:H\to H$ is nontrivial for some $k\in K$, so $\phi_k(h)\ne h$ for some $h\in H$, in which case $khk^{-1}=\phi_k(h)\ne h~\,\Rightarrow~ kh\ne hk$ within $H\rtimes K$.

1

$H$ and $K$ are subgroups of $H\rtimes K$, so if $H\rtimes K$ then $H$ and $K$ had better be abelian too. The homomorphism $\varphi:K\to\operatorname{Aut}(H)$ becomes conjugation in $H\rtimes_\varphi K$. Conjugation is always trivial in abelian groups, so again, if $H\rtimes_\varphi K$ is abelian, then $\varphi$ had better be trivial. In fact, the ...

1

First, +1 for mentioning that this is homework. I'll stick to providing hints and let you fill in the gaps, since you seem to have a grasp of what to do. For $\it{a.}$, if you rearrange what you've already worked out, then we see that the coset of $H$ containing $(m,n)$ depends only on the difference between $m$ and $n$. Can you use this to write down a ...

1

Hint: An element in $\;S_n\;$ has order $\;q=$ a prime iff it is the product of disjoint $\;q-$cycles. Now, in $\;S_p\;$ how many disjoint $\;p-$cycles are there?

1

The elements which satisfy the above equation are only $p$ cycles and product of disjoint $p$ cycles, but there is no permutation which is product of disjoint $p$ cycles and there are $(p-1)!$ $p$-cycles, so there are $1+(p-1)!$ elements satisfying the equation as identity also satisfy the equation.

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