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6

It is always true that $z\overline z = |z|^2$, so when $|z| = 1$, we may conclude that $\overline z = 1/z$. edit: and you know that roots of unity have $|z| = 1$ because $|z^k| = |z|^k$ (which you can prove by showing $|wz| = |w||z|$ and repeating) and hence any $n^\mathrm{th}$ root of unity has $|z|^n = 1$, which (since $|z|$ is a positive real number) ...


6

$\mathbb R/\mathbb Z$ identifies each point on the real line with a unique point in the interval $[0,1)$. We can stretch this interval by a factor of $2\pi$ and wrap it around the circumference of the unit circle in the complex plane. Because the real number $1$ maps to $0$ in the interval $[0,1)$ we find that joining the ends together works, and each real ...


4

Here's another way of looking at this through a reverse lens: Let's solve $u^3+3u-4=0$ by Cardano's method, putting $u=x+y$. Then $(x+y)^3-3xy(x+y)-(x^3+y^3)=0$ and we require: $$x^3+y^3=4$$ and $$-3xy = 3 \text { so that }xy=x^3y^3=-1$$ Then we note that $x^3$ and $y^3$ are roots of the quadratic $$z^2-4z-1=0$$So that $$z=\frac{4\pm\sqrt{16+4}}{2}=2\pm ...


3

Well, $\;R:=\Bbb Z/p^n\Bbb Z\;,\;\;n>1\;$ cannot be a vector space over $\;\Bbb F_p\;$ since, for example $$\overline1:=1+p^n\Bbb Z\;\;:\;\;0\neq \underbrace{\overline 1+\overline1+\ldots+\overline1}_{p\;\text{times}}=p\cdot(1+p^n\Bbb Z)=0$$ since $\;p=0\;$ in $\;\Bbb F_p\;$


3

From a bottom-up perspective, one thing we do with mathematical structures is combine them to make new ones. The most basic way of doing this with e.g. groups and rings is the direct sum. This is how get get vector spaces from scalar fields, after all (up to isomorphism), using coordinates. In the top-down perspective, one thing we do with mathematical ...


2

To show that $\mathbb{Z}/(p^n)$ does not contain a field, suppose to the contrary that it contains the field $K$. Think of $\mathbb{Z}/(p^n)$ as the integers from $0$ to $p^n-1$, under the usual addition and multiplication modulo $p^n$. Let $a$ be a non-zero element of $K$. Note that $a$ cannot be divisible by $p$, else a power of $a$ would be the ...


2

As for vector spaces, to show that a given set $S$ generates the group, it suffices to show that the elements of a known generating set can be expressed in term of $S$. In this case you can take the standard generators $(1,0)$ and $(0,1)$ of $\Bbb Z\times\Bbb Z$ and express them as integer combinations of the vectors given. While a general method to find ...


2

No, it is not enough. There could be many more cosets. Consider for instance the Klein four-group $G$, and $H = \{1\}$. Spoiler


2

The comments suggest that you mean only centralizers of involutions. Even in that case, no finite group $G$ can have only the two involutionn centralizers you suggest, so Q2 seems to have a negative (or empty) answer. One involution centralizer must contain a Sylow $2$-subgroup. Hence the Sylow $2$-subgroup of $G$ must have order $16$. But the elementary ...


2

I am not 100% sure that this is what you want, but I think that $$\begin{align} \Big[-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y},\frac{\partial}{\partial x}\Big] &= \left(-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}\right)\frac{\partial}{\partial x} - \frac{\partial}{\partial x}\left(-y\frac{\partial}{\partial ...


2

Consider the Koszul resolution $K_\bullet$ of $A=A_1(k)$ as an $A$-bimodule —this was probably first constructed in [Sridharan, R. Filtered algebras and representations of Lie algebras. Trans. Amer. Math. Soc. 100 1961 530--550. MR0130900 (24 #A754)] (which you can find here); you can find a description of the complex here. Since $K_\bullet$ is of finite ...


2

A quick answer.. For a fixed positive integer $n$, An equivalence relation ~ on $\mathbb Z$, The ring of Integers, defined by $a$~$b$ $\Leftrightarrow n$ devides $a-b$ , partitions $\mathbb Z$ into $n$ equivalence classes $\bar 0, \bar 1, \cdots ,\overline{n-1}$ $\quad$ where $\bar r=\{ nk+r: k\in \mathbb Z\}$, $0\leq r\leq n-1$. The quotient set $\mathbb ...


2

The theorem you haven't proved states that any f.g. abelian group is isomorphic with $$\Bbb Z^n\oplus T\;,\;\;\text{with the torsion part $\;T\;$ being the subgroup of all elements of finite order}$$ Thus, a f.g. abelian group is finite iff $\;0=n=$ the free rank of the group.


2

Let $H\subseteq G$ be the unique non-trivial proper subgroup of $G$. Since $H$ is a proper subgroup there is some $g\in G\setminus H$. The subgroup $\langle g\rangle$ generated by $G$ is distinct from $H$ since $g\notin H$ and therefore can not be a proper subgroup. It follows that $G=\langle g\rangle$ for every $g\notin H$, in particular $G$ is cyclic. ...


2

Because the homomorphism $$ \varphi\colon\mathbb{R}\to \mathbb{C}\setminus\{0\} $$ (the domain is a group with respect to addition, the codomain with respect to multiplication) defined by $$ \varphi(t)=\cos(2\pi t)+i\sin(2\pi t) $$ has the unit circle as its image and $\mathbb{Z}$ as its kernel.


2

It's pretty clear that every pair of distinct elements with order 2 will generate such a subgroup. Seek all of those pairs! If I didn't miscount, I think I see 7 elements of order 2. The subgroup will contain three elements of order 2. Experiment and see what combinations are distinct. Be warned though: it won't be this simple for higher orders and ...


2

If $G$ is a group with operation $*$, that $g$ is a generator means that every element can be written as $$ g *g * \cdots * g $$ or is the inverse of such a thing, or is the identity. When the group operation is written as multiplication, we abbreviate $$ \underbrace{g *g * \cdots * g}_{n\;g\text{s}} = \underbrace{gg\cdots g}_{n\;g\text{s}} = g^n \qquad ...


2

They key part is recognising part (c). This is false. We know that there is only $1$ Sylow 11-subgroup of $G$. And so this must be a normal subgroup of $G$. Call it $H$. Then consider $G/H$. This has order 4. As such is must be isomorphic to either $\mathbb{Z}_2\times \mathbb{Z}_2$ or $\mathbb{Z}_4$. Hence $G \cong H \times \mathbb{Z}_4$ or $G \cong H ...


2

Suppose $G$ is a noneabelian group of order $6$. By Cauchy's theorem, there exists an element of order $2$, call it $b$, and an element of order $3$, call it $a$. The subgroup $H=\langle a\rangle$ is normal in $G$, since it has index $2$. Since $\langle a \rangle\cap\langle b\rangle =1$, $\langle a,b\rangle =G$. We have $bab^{-1}=a^i$, it cannot be $i=1$ ...


1

There are lots of examples. Many Frobenius groups have this property (that includes the dihedral groups of twice odd order, and other examples such as nonabelian groups of order $pq$ for primes $p,q$). $A_4$, $A_5$, ${\rm SL}(2,3)$ and ${\rm SL}(2,5)$ are examples. It seems likely that $A_5$ is the only nonabelian simple group with this property, and it ...


1

The Cayley table: $$\begin{array}{c|ccc} \ast & 0 & 1 & 2 \\ \hline 0 & 0 & 2 & 1 \\ 1 & 1 & 0 & 2 \\ 2 & 2 & 1 & 0 \end{array}$$ represents a finite quasigroup of order $3$ on the set $\mathbb{Z}_3$ of the integers mod$3$. The operation $\ast$ is $$a\ast b=(a-b)\text{mod}3.$$ Check that the operations ...


1

If $G$ were not cyclic, then there must be two distinct elements $a$ and $b$ that generate two distinct cyclic subgroups. Thus, $G$ must be cyclic. From here, we know all cyclic groups are isomorphic to $\mathbb{Z}_n$, where $n = |G|$. At this point, we are done: For all cyclic groups $H$, there will exist exactly $1$ cyclic subgroup for each divisor of ...


1

The standard (positive) division algorithm asserts that $0\leq u_1<n$. There is a second Euclidean algorithm which allows the remainder, $u_1$, to be negative - where then the condition is $\left|u_1\right|\leq \frac{n}{2}$. This needs to be proved, but it is an easy corollary of the positive division algorithm. For $a=9,n=5$, you'd have $u=2$ and ...


1

As $hxh^{-1} \in X$, the map $\phi_h : X \rightarrow G$ defined by $\phi_h(x^k)=hx^kh^{-1}$ defines an automorphism of $X$. In fact, there is a map $\phi : H \rightarrow \mathrm{Aut}(X)$ given by $\phi(h^k)=\phi_h^k$. Now, since $X$ is cyclic of prime order, the automorphisms of $X$ are specified uniquely by sending our particular generator, $x$, to one of ...


1

Since $H\lhd G$, we can consider the canonical homomorphism $G\to G/H$. Under this, $x$ is mapped to an element of $G/H$ of order dividing $11$ (i.e. $1$ or $11$). Since $G/H$ has order $12$, there is no element of order $11$, hence $x$ maps to an element of order $1$, i.e. the neutral element of $G/H$. This is equivalent to $x\in H$.


1

You can for example show that the algebraic closure of $\mathbb{Q}$ contains arbitrarily large $\mathbb{Q}$-linearly independent set. Pick your favourite one, mine would probably be the roots of $x^n-2$ (which is irreducible, by Eisenstein criterion, for every possible choice of $n$).


1

Linear independence in an Abelian group should be an example of linear independence in a module. Each Abelian Group is a module over the integers, since we can multiply an element by a positive integer by adding it to itself that many times, and by a negative integer by adding the inverse to itself that many times. The definition of linear independence for a ...


1

The space consists of just nine points $(0,0)$, $(0,1)$, $(0,2)$, $(1,0)$, $(1,1)$, $(1,2)$, $(2,0)$, $(2,1)$, $(2,2)$ which can be naturally put on a $3\times3$ grid. The pictures show arrows which connect each point to the point it is mapped to (the latter being where the arrow points to), with the arrows for points mapping to itself omitted (thus the ...


1

This isn't exactly an answer to your question, but shows how to put this example into a more general context. You can use two more general results, which prove that a group $G$ of order $p^2$ is always abelian when $p$ is a prime. The centre $Z$ of $G$ is non-trivial. Here one uses the fact that the sizes of the conjugacy classes in $G$ divide the order ...



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