## abstract algebra – Let $phi: mathbb {Z} _7 times mathbb {Z} _7$ be a homomorphism such that $phi ^ 5 = text {id}$. Show that $phi$ is the identity.

Let $$phi: mathbb {Z} _7 times mathbb {Z} _7 to mathbb {Z} _7 times mathbb {Z} _7$$ to be a homomorphism such as $$phi ^ 5 = text {id}$$. CA watch $$phi$$ is the identity.

My attempt: Since $$ker ( phi) subset ker ( phi ^ 2) subset cdots subset ker ( phi ^ 5) = 0$$it follows that $$phi$$ is injective. Since $$| mathbb {Z} times mathbb {Z} |$$ is finished, $$phi$$ is automatically an isomorphism.

then $$phi$$ should map a subgroup to a subgroup. Consider $$phi ( mathbb {Z} _7 times 0)$$, so he should be associated with $$mathbb {Z} _7 times 0$$ or $$0 times mathbb {Z} _7$$. Without loss of generality, we consider the first case. In particular, $$phi ( 1.0)$$ should be a generator for $$mathbb {Z} _7 times 0$$. assume $$phi (1,0) = (a, 0)$$. then $$(1,0) = phi ^ 5 (1,0) = phi ^ 4 (a, 0) = phi ^ 3 (a ^ 2,0) = cdots = (a ^ 5,0)$$. So, $$a ^ 5 = 7k + 1$$ for some people $$k in mathbb {Z}$$. By testing $$a in mathbb {Z} _7$$, we see that the only possibility is that $$a = 1$$. The discussion for $$phi (0,1)$$ Is similar.

Could someone help me take a look at my attempt? Is my approach reasonable and is there a better way to solve this problem?

## Abstract algebra – How to apply the theory of "homology" to the Goldbach conjecture using the result of Helfgott (ternary Goldbach true)?

The result of Helfgott: every odd integer $$n geq 7$$ is the sum of $$3$$ (odd) bonuses. I saw somewhere in the newspaper the mention of odd prime numbers. It means that every single integer $$n geq 10$$ is the sum of 4 odd prime numbers (you just added a $$3$$ to each odd number).

Let's first assume something bold, to simplify the discussion, that for each integer $$x geq 13$$, the even number $$2x$$ can be expressed as the sum of $$4$$ distinct prime numbers: $$2x = p + q + r + s$$. To handle all cases, I'm sure it's possible, because it would take either an enumeration of the small set of cases, or a combinatorial argument. Reason is $$x geq 13$$ is because: $$3 + 5 + 7 + 11 = 26 = 2 cdot 13$$ is the smallest even number which is the sum of $$4$$ distinct odd primes.

If you are not familiar with the basics of simplicial homology but you know what the free $$Bbb {Z}$$-module on a set of symbols is, then google "introduction to simplicial homology". You basically need limit cards: $$partial_n: C_n to C_ {n-1}$$ who take each element of the module $$C_n$$ at its "limit" in $$C_ {n-1}$$, its border being represented as a formal element in the free module, whatever the border map, it must be a $$Bbb {Z}$$homomorphism -module and satisfy also: $$partial_ {n-1} circ partial_n = 0$$.

So, according to Helfgott's result (and the net distinction hypothesis), each integer $$x geq 13$$ is such that the even number $$2x = p + q + r + s$$, or $${p, q, r, s }$$ are distinct odd prime numbers. As a result, there is $${4 choose 2} = 6$$ ways of forming even smaller numbers (by adding two of the prime numbers involved in the sum of $$x$$) $$2x_i, i in {1, dots, 6 }$$ such as $$2x_i + 2x_j = 2x$$ or equivalent $$x_i + x_j = x$$. Now, there is another problem with duplicates as $$x_i = x_j$$ is a possibility. But for now, suppose that $$x_i neq x_j$$, for reasons of discussion. So, to recap, everyone $$x_i = u + v$$ or $$u, v in {p, q, r, s }$$.

So, if we associate each $$x_i$$ with the official symbol $$bar {x_i}$$ which might not be an integer (for the moment) then we will call $$bar {x_i}$$ the faces of the simplex $$bar {x}$$. But $$bar {x}$$ could also be a simplex face of another integer! It is therefore possible to obtain a long-chain complex eventually.

So we have at least $$Delta = { bar {x}, bar {x_i}: i in {1, points, 6 } }$$ like a complex. This does not necessarily correspond to an abstract simplicial complex in which $$forall sigma in Delta,$$ if $$tau subset sigma$$then $$tau in Delta$$. Since for one, we have not defined the $$bar {x_i}$$ to be sets – these are just formal symbols.

Another approach would be?

remember that $$sum limits_ {i = 1} ^ 6 alpha_i x_i = x$$ or $$sum limits_ {i = 1} ^ 6 alpha_i = 2, alpha_i in {0, 1$$. It's a bit like a "convex combination". Anway, as you can see, I took elements of simplicial homology and tried to adapt it. Do you have any ideas on how to proceed?

## abstract algebra – $I$ is a free R-module if and only if $I = Ra$

Problem: Let $$I$$ to be a non-zero ideal of a commutative ring $$R$$ with identity. Prove it
$$I$$ is a free R-module if and only if $$I = Ra$$ for some people $$a in R$$ it's not a zero
divider.

My idea is:
($$Rightarrow$$) Let $$I$$ to be generated by $${a_1, … a_n } (n> 1)$$, then I have $$a_1x_1 + … + a_nx_n = 0$$with $$x_1 = a_2, x_2 = -a_1, …$$($$n$$ is same); $$x_1 = a_2, x_2 = -a_1-a_3, x_3 = a_2, x_4 = a_5, x_5 = -a_4, …$$ ($$n$$ is odd).
So, $$a_1 = … = a_n = 0$$. Contradiction. So, $$n = 1$$.

Is it good? Help me! Thank you!

## abstract algebra – The finite Boolean ring with $1 neq 0$ is isomorphic to $mathbb {Z} _2 times mathbb {Z} _2 times dots times mathbb {Z} _2$

Let $$R$$ to be a finite boolean ring with $$1 neq 0$$. CA watch $$R cong mathbb {Z} _2 times mathbb {Z} _2 times dots times mathbb {Z} _2$$.

This is Exercise 2 on p267 of the book Dummit and Foote's Abstract Algebra. The author gives the idea of ​​using it if $$e$$ is a idempotent of a ring $$R$$ with $$1$$then

$$Re times R (1-e)$$

or $$Re$$ is a ring of identity $$e$$ and $$R (1-e)$$ an identity ring $$1-e$$.

My attempt:

Yes $$R = {0,1 }$$then $$R cong mathbb {Z} _2$$ is trivial.

Therefore, we can assume that $${0,1 }$$ is properly contained in $$R$$. So, we can choose an element $$x notin {0,1 }$$. We get because every element of a boolean ring is a idempotent:

$$Rx times R (1-x)$$

$$Rx$$ is a ring of identity $$x neq 0$$ and $$R (1-x)$$ is a ring of identity $$1-x neq 0$$. In addition, $$Rx$$ and $$R (1-e)$$ have a cardinality strictly inferior to the cardinality of $$R$$. These rings are also Boolean. The induction gives the result. $$quad square$$

Is it correct?

## abstract data types – Where is the original literature for the specification of an associative array ADA?

I am looking for the formal specification of abstract data type algebra for associative array, associative list (AKA multimap) and other similar ADT to quote for use in a data specification similar to binary JSON . The Wikipedia associative table entry uses non-standard mathematical terminology and their sources are all secondary sources without original sources. There is an algebra for the dictionary. It starts with Let x be a set and defines the operations on this set. Who invented the dictionary and when? It should be quite old, possible from the Journal of the Association of Computer Machinery or a similar publication.

## abstract algebra – My question is about the normal subgroup of dihedral groups

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## abstract algebra – How is Galois theory applied here?

I have not studied galois theory much and it's been a long time, so I could really use a sentence in this article about an algorithm for polynomial factorization.

Preliminaries are like in another question of mine to find it here:

• $$f in Z (X)$$ monic, squarefree (meaning here no multiple roots) with degree $$N$$
• A prime number $$p$$ such as $$f mod p$$ stay squarefree (I do not know if it's important here)
• $$f = prod_ {i = 1} ^ n f_i$$ in $$mathbb {Z} _p (X)$$

We also define (for a field $$F$$ and some $$i in mathbb {N}$$) the
$$i$$-trace of a polynomial $$g in F (X)$$ as $$Tr_i (g): = > sum_ {l = 1} ^ { deg (g)} zeta_l ^ i$$ where the $$zeta_l$$ are the
$$deg (g)$$ (not necessarily different) roots of $$g$$, as well as, for a
full $$l$$ $$Tr_ {1 dots l} (g): = left (Tr_1 (g), , dots, Tr_l (g) right)$$ Since $$Tr_i (gh) = Tr_i (g) + Tr_i (h)$$, all these terms can also be
defined more generally for $$g in F (X)$$ if we put $$Tr_i (q / r) = Tr_i (q) – Tr_i (R)$$.

Also for $$i in {1, , dots, n }$$, to put
$$V_i: = Tr_ {1 dots N} (f_i)$$
and assume $$V_i in mathbb {Q_p} ^ N$$ for everyone $$i$$.

Then if $$g = prod_ {i = 1} ^ n f_i ^ {v_i}$$ for arbitrary integers $$v_i$$, $g in mathbb {Q_p} (X)} and using the extended definitions of the trace implied above, $$Tr_ {1 , points N} (g) = V: = sum_ {i = 1} ^ n V_i$$ Now, I already know that the $$V_i$$ in the case where I look are linearly independent, so we have an isomorphism of groups $$Tr_ {1 dots N}: left ( left { prod_ {i = 1} ^ n f_i ^ {v_i}; ; v_1, , dots, v_n in mathbb {Z} right }, , cdot right) rightarrow left ( mathbb {Z} V_1 + , points + mathbb {Z} V_n, , + right) subset mathbb {Q} _p ^ N$$ Suppose, moreover, $$V in mathbb {Z} ^ N$$. Now, the statement that I do not understand is The group of Galois on $$mathbb {Q}$$ leaves $$V$$ invariant, so it permutes the preforms of $$V$$, so the fact that the card is 1-1 implies that $$g$$ is invariant and therefore defined on $$mathbb {Q}$$. I do not really see where exactly Galois theory is applied here … are we talking about the Galois group of the Galois extension $$mathbb {Q} ( alpha_1, , points, , alpha_n) / mathbb {Q}$$? How / where is it applied and what is the link with the previous images, etc.? I do not really know where to look / what to look for, because the sentence is written for people who already understand exactly what it's talking about here, and I do not … ## abstract algebra – theory of representation notation doubt I've therefore read a little representation theory in the book "Linear representations of finite groups" and I have a question. When the author writes: Let n be the order of G, and V is a vector space of dimension n, with a base $$(e_t) _ {t in G}$$ indexed by the elements t of G. "I do not know how to look at it, I do not think it's supposed to be $$mathbb {R} ^ n$$ I guess my doubt is that I do not understand what he means with this basic notation. So any help is appreciated thanks. ## abstract algebra –$ R $commutative,$ I $maximal w.r.t. ideals not generated, show that$ I \$ is not generated.

Let $$R$$ be a commutative ring with the unit, with at least one ideal that is not generated finitely. Let $$S$$ to be the family of such ideals. Show $$S$$ has a maximum element.

I need to show every channel $$S$$ has an upper bound and uses Zorn's lemma. I can take a chain $$mathcal {C}: I_ {1} subset I_ {2} subset I_ {3} subset cdots$$ in $$S$$ and let $$I: = bigcup_ {n} I_ {n}$$. Yes $$I = (a_ {1}, ldots, a_ {t})$$ is finished generated, so I want to come to a contradiction. From here I can say that $$I_ {n} = (b_ {n, 1}, b_ {n, 2}, ldots) subset (a_ {1}, ldots, a_ {t})$$ for each $$n in mathbb {N}$$, from where $$b_ {n, i} = r_ {n, 1} ^ {i} a_ {1} + cdots + r_ {n, t} ^ {i} a_ {t}$$ (a linear combination). However, I do not know where to go from there, or if it's even the right direction. I guess I should use the inclusion $$I_ {1} subset I_ {2} subset cdots$$ In a way, seeing as subidels of ideals finely generated does not need to be finely generated.

## Abstract Algebra – Proof of Sylow's Theorem

I read Introduction to Alexander Paulin's Abstract Algebra, and I could find a line in the proof of Sylow's Theorem.

It's on the 24th and 25th page of the notes, quoted below, where the line I do not get is highlighted

Now remember that $$S$$ is the disjoint union of the orbits of our action
of $$G$$ sure $$S$$.

(On the notation: $$Stab ( omega)$$ is the stabilizing subgroup; $$Orb (omega)$$ is the orbit; and HCF means the highest common factor):

Sylow's theorem. Let $$(G, *)$$ to be a finite group such as $$p ^ n$$ Split $$| G |$$, or $$p$$ is first.
Then there is a subgroup of order $$p ^ n$$.

Evidence. Suppose that $$| G | = p ^ nm$$, or $$m = p ^ ru$$ with $$HCF (p, u) = 1$$. Our central strategy
is to consider an intelligently chosen group action of G and to prove one of the subgroups of stabilizers
to the size $$p ^ n$$. We will need to heavily exploit the orbit stabilizer theorem.

Let $$S$$ be the set of all subsets of $$G$$ size $$p ^ n$$. An element of $$S$$ is an unordered $$n$$-tuple
of distinct elements in $$G$$. There is a natural action of $$G$$ sure $$S$$ by term term composition
on the left.

Let $$∈ S$$. If we correct an order $$ω = {ω_1, · · ·, _ {p ^ n}} S$$then $$g (ω): = {g * _1, · · ·, g * _ {p ^ n}}$$.

• We claim first that $$| Stab (ω) | ≤ p ^ n$$. To see this define the function
$$f: Stab (ω) → ω$$
$$g → g * _1$$
By the property of cancellation for groups it is an injective card. Therefore $$| Stab (ω) | ≤ | ω | = p ^ n$$.

• Observe this
$$| S | = pmatrix {p ^ nm \ p ^ n} = frac {(p ^ nm)!} {p ^ n! (p ^ mp ^ n)!} = prod_ {j = 0} ^ {p ^ n-1} frac {p ^ nm-j} {p ^ nj} = m prod_ {j = 1} ^ { p ^ n-1} frac {p ^ mj} {p ^ nj}$$

Note that if $$1 ≤ j ≤ p ^ n – 1$$ then j is divisible by $$p$$ at most $$n – 1$$ time. It means
this $$p ^ nm – j$$ and $$p ^ n – j$$ have the same number of $$p$$ factors, namely the number of $$p$$
factor of $$j$$. It means that
$$m prod_ {j = 1} ^ {p ^ n-1} frac {p ^ m-j} {p ^ n-j}$$

do not have $$p$$ factors. Therefore $$| S | = p ^ rv$$, or $$HCF (p, v) = 1$$.

## Now remember that $$S$$ is the disjoint union of the orbits of our action of $$G$$ sure $$S$$.

Therefore
there must be a $$∈ S$$ such as $$| Orb (ω) | = p ^ st$$, or $$s ≤ r$$ and $$HCF (p, t) = 1$$.

By
the orbiting theorem we know that $$| Stab (ω) | = p ^ {n + r – s} frac {u} {t}$$.

Because $$| Stab (ω) | ∈ N$$
and $$u$$ and $$t$$ are coprime to $$p$$, we can deduce $$frac {u} {t} N$$. Therefore $$| Stab (ω) | ≥ p ^ n$$.

• For this choice of S, Stab () is a subgroup of size pn.

So I do not understand this line:

Now remember that $$S$$ is the disjoint union of the orbits of our action
of $$G$$ sure $$S$$.

What does it mean? for example, if take $$G$$ as $$Sym_3 = {e, a, b, c, d, f }$$, the transformers of a triangle, where $$a$$, $$b$$ are counterclockwise and clockwise rotation; $$c$$,$$d$$,and $$f$$ like reflections. Let $$p = 3$$, $$n = 1$$, $$m = 2$$.

then $$S$$ would be the set of all subsets of $$Sym_3$$ size $$3$$. Picking $$3$$ elements of $$6$$ at $$pmatrix {6 \ 3} = 20$$ choice, so $$S$$ at the size of $$20$$to know
$$S = {{{e, a, b }, {e, a, c }, {e, a, d }, {e, a, f }, {e, b, c }, {e, b, d }, {e, b, f }, {e, c, d }, {e, c, f }, {e, d, f , {a, b, c }, {a, b, d }, {a, b, f }, {a, c, d }, {a, c, f }, {a, d, f }, {b, c, d }, {b, c, f }, {b, d, f }, {c, d, f }$$

So, what disjointed orbits formed this $$S$$?