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4

A different approach. Assume that $i$ is such that $\alpha^i\neq1$. Let $$ \sigma_i=\sum_{x\in\Bbb{F}_{p^n}}x^i $$ be the sum. We get that $$ \alpha^i\sigma_i=\alpha^i\left(\sum_{x\in\Bbb{F}_{p^n}}x^i\right) =\sum_{x\in\Bbb{F}_{p^n}}\alpha^ix^i= \sum_{x\in\Bbb{F}_{p^n}}(\alpha x)^i. $$ Here $\alpha x$ ranges over the field when $x$ does, so the last sum is ...


4

I will try and generalize Nicky Hekster's solution to the coprime case. Let $\gcd(m,n)=r$, with $m=ra$, $n=rb$, and $a,b$ coprime Let $S = \{g^r \mid g \in G \}$, and let $M = G_m = \langle k^a \mid k \in S \rangle$, and $N = G_n = \langle k^b \mid k \in S \rangle$. So $M$ and $N$ are abelian normal subgroups of $G$, with $[M,N] \le M \cap N \le Z(\langle ...


3

Think of $\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. It has order $4$ but all elements have order one or two.


3

Hint $(\Bbb Q_{> 0}, \cdot)$ is isomorphic to $(\Bbb Z, +) \oplus (\Bbb Z, +) \oplus \cdots$.


3

The smallest $n$ is $6$: 1. $A$ is isomorphic to $\langle(1,2),(3,4),(5,6)\rangle$. 2. For $n=4,5$ the only subgroup of order $8$ which $S_n$ does contain is the dihedral group $D_4$ (and its conjugates, being a $2$-Sylow subgroup).


3

Consider the dihedral group $D_6 = \langle x, y\mid x^3 = 1, y^2 = 1, yx = x^2y\rangle$. $K = \{1,y\}$ is a subgroup which isn't normal. $xKx^{-1} = \{1,xyx^{-1} = xyx^2 = xx^2yx = xx^2x^2y = x^5 y = x^2y\} \not= K$.


2

For any finite subset $G\subset \mathbb N$ and map $\circ\colon G\times G\to \mathbb N$, we can find a polynomial $f\in\mathbb Q[X,Y]$ such that $f(a,b)=a\circ b$ for all $a,b\in G$.


2

The fundamental theorem of finite Abelian groups says that every finite Abelian group is a direct product of cyclic groups of prime power order. Can you use that to come up with a proof?


2

By the second isomorphism theorem, the condition that $HK=G$, or probably in your case $H+K=G$ (if we're using additive notation) is sufficient.


2

It's not true for $P=\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/8\mathbb{Z}$, is it? The subgroup $Q$ generated by $(1,2)$ is characteristic (it's the only cyclic $Q$ of order $4$ such that $P/Q$ is also cyclic), but it's not of the form you describe. Although it's not immediately obvious how to give a similar example for odd $p$.


2

Group of order 1 is trivial, groups of order 2,3,5 are cyclic by lagrange theorem so they are abelian. For a group of order 4, if it has an element of order 4, it is abelian since it is cyclic(isomorphic to Z4). If orders of every element are 2, then the inverse of an element is the element itself so you can verify every element commute with each other so ...


2

Yes, it is possible to prove. The question is, how much group theory you can use. Any group of prime order is cyclic, hence abelian. This implies that all groups of order $2$, $3$ and $5$ are abelian. On the other hand, all groups of order $p^2$ are abelian. Hence all groups of order $4$ are abelian. The trivial group of order $1$ is abelian as well. For ...


2

$xy \neq yx \implies G$ has at least five elements: $e,x,y,xy,yx$. But $|G| = 5$ means that $G$ is isomorphic to $\mathbb{Z}_5$, so every group of order less than 6 must be abelian.


1

Another way to prove it is that a finite ordered non-abelian group must have order 6 or greater which can be done like this if $|G|=1$ then it's trivial, So assume $|G|>1$ then there exist $a\in G$, but all groups have inverses so $a^{-1}\in G$, which means $|G|\geq 3$, but this case would be abelian so there must be another element $b\in G$ and equally ...


1

Order of a group , $n$ starts from 1. When $n=1$ the group is a trivial one. Now every group of prime order is cyclic and hence abelian. Hence groups of $n=2,3$ and $5$ are abelian. Since every group of order $\ p^2$ (where $p$ is prime) is abelian. Group of order $4=\ 2^2$ is abelian. Hence every group of order less than or equal to $5$ is ...


1

If you can use the structure theorems for finite abelian groups, then it is easy: In the invariant factor decomposition, the largest factor has order equal to the exponent of $G$. Any $a\in G$ whose order is the exponent of $G$ must be a generator of this factor. If you want to start with the primary decomposition, then the Chinese remainder theorem allows ...


1

If $G\neq 1$, then $G$ being locally graded, it has a proper subgroup of finite index $H$. Hence $H$ is abelian; in particular $G$ is virtually abelian. Assume that $G$ is not abelian and infinite. Then any two non-commuting elements of $G$ generate $G$; in particular $G$ is finitely generated. Being finitely generated and virtually abelian, $G$ has a ...


1

The group $\langle x,y | xy=yx, x^5=y^3 \rangle$ is the quotient of $\langle x,y | xy=yx \rangle = \mathbb{Z} \oplus \mathbb{Z}$ (with $x=(1,0)$ and $y=(0,1)$) by the (normal) subgroup generated by $5x-3y=(5,-3)$, and similarly $\langle x,y | xy=yx , x^4=y^2 \rangle$ is the quotient of $\mathbb{Z} \oplus \mathbb{Z}$ by the (normal) subgroup generated by ...


1

It's easier to first talk about endomorphisms because these have a ring structure. In fact, as has been mentioned a few times already, this group is a countable direct sum $\bigoplus_p \mathbb{Z}$ of a copy of $\mathbb{Z}$ for every prime, and hence in some respects it behaves like a vector space. In particular, its endomorphism ring $\text{End}(\bigoplus_p ...


1

This is one of those places where the details hide the picture. Suppose $G$ is a group (not necessarily abelian), with $A$ and $B$ subgroups of $G$. We can define $$ AB=\{ab:a\in A,b\in B\}=\bigcup_{a\in A} aB = \bigcup_{b\in B} Ab $$ If $B$ is a normal subgroup of $G$, then $AB$ is a subgroup of $G$ Indeed, $1$ obviously belongs to $B=1B\subseteq ...


1

Since $G$ is abelian, the map $\prod H_i \to G$ defined by $(h_1, \dots, h_k) \to h_1 \cdots h_k$ is a homomorphism. It's not hard to prove by induction on $k$ that this map is injective. (In brief: If $h_1 \cdots h_k = 1$, then $h_1^{\alpha_k} \cdots h_{k-1}^{\alpha_k} = 1$, and the order of each $h_i$ in the product is prime to $\alpha_k$.) Since $\# \prod ...



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