### '★MATH/2. 현대대수'에 해당되는 글 12건

1. 2012.07.13 Eisenstein 판별법 (1)
2. 2010.04.07 finite simple group
3. 2010.03.28 정12면체군
4. 2010.03.25 even & odd permutation
5. 2010.03.24 정다면체
6. 2010.03.22 군표현론 & homomorphism
7. 2010.03.17 cydlic group 2
8. 2010.03.17 cyclic group
9. 2010.03.15 순환군
10. 2010.03.06 현대대수 솔루션

## Eisenstein 판별법

2012.07.13 15:48
어떤 정수계수 다항식이 유리수 범위 안 에서 인수분해가 가능한지 판별하는 방 법입니다. 모든 다항식에 적용하는건 아니고요. 이 러이러한 조건을 만족해야 됩니다...

임의의 정수계수 다항식을 생각합니다... 어떠한 소수 p∈Z 가 존재해서 다음을 만 족해야 됩니다.

1. 최고차항은 p의 배수가 아니다.
2. 최고차항 제외한 나머지 모든항의 계 수는 p의 배수이다.
3. 상수항은 p의 배수이지만, p² 의 배수 는 아니다.

이러한 조건을 만족하면 이 다항식은 유 리수 범위 안에서 인수분해가 불가능 합 니다. 이러한 판별 방법을 Eisenstein 판별법 이 라 합니다.

책에 간단한 예제를 풀어보면...

f(x)=25x^5 -9x^4 -3x^2 -12

이거는 p=3 이라는 숫자를 생각하면

1. 최고차항은 3 의 배수가 아니죠
2. 최고차항을 제외한 모든항은 3의 배수
3. 단 상수항은 3의 배수이지만 9의 배수 는 아니죠.

위와 같은 조건을 만족하는 p 이라는 숫 자를 찾을수 있으면 위의 다항식은 유리수 범위 안에서 인수 분해가 불가능 합니다.

출처@
http://m.kin.naver.com/mobile/qna/detail.nhn?d1id=11&dirId=1113&docId=67301999&qb=ZWlzZW5zdGVpbg==&enc=utf8&section=kin&rank=1&search_sort=0&spq=0

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 Eisenstein 판별법  (1) 2012.07.13 2010.04.07 2010.03.28 2010.03.25 2010.03.24 2010.03.22

## finite simple group

2010.04.07 01:49

http://en.wikipedia.org/wiki/List_of_finite_simple_groups

http://en.wikipedia.org/wiki/Simple_group#Finite_simple_groups

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 Eisenstein 판별법  (1) 2012.07.13 2010.04.07 2010.03.28 2010.03.25 2010.03.24 2010.03.22

## 정12면체군

2010.03.28 12:03

http://ko.wikipedia.org/wiki/%EC%A0%95%EC%9D%B4%EB%A9%B4%EC%B2%B4%EA%B5%B0

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 Eisenstein 판별법  (1) 2012.07.13 2010.04.07 2010.03.28 2010.03.25 2010.03.24 2010.03.22

## even & odd permutation

2010.03.25 00:47

http://en.wikipedia.org/wiki/Parity_of_a_permutation#Equivalence_of_the_two_definitions

even permutation

An even permutation is a permutation obtainable from an even number of two-element swaps, i.e., a permutation with permutation symbol equal to . For initial set 1,2,3,4 , the twelve even permutations are those with zero swaps: ( 1,2,3,4 ); and those with two swaps: ( 1,3,4,2 , 1,4,2,3 , 2,1,4,3 , 2,3,1,4 , 2,4,3,1 , 3,1,2,4 , 3,2,4,1 , 3,4,1,2 , 4,1,3,2 , 4,2,1,3 , 4,3,2,1 ).

For a set of elements and , there are even permutations, which is the same as the number of odd permutations. For , 2, ..., the numbers are given by 0, 1, 3, 12, 60, 360, 2520, 20160, 181440, ... (Sloane's A001710).

odd permutaition

An odd permutation is a permutation obtainable from an odd number of two-element swaps, i.e., a permutation with permutation symbol equal to . For initial set 1,2,3,4 , the twelve odd permutations are those with one swap ( 1,2,4,3 , 1,3,2,4 , 1,4,3,2 , 2,1,3,4 , 3,2,1,4 , 4,2,3,1 ) and those with three swaps ( 2,3,4,1 , 2,4,1,3 , 3,1,4,2 , 3,4,2,1 , 4,1,2,3 , 4,3,1,2 ).

For a set of elements and , there are odd permutations (D'Angelo and West 2000, p. 111), which is the same as the number of even permutations. For , 2, ..., the numbers are given by 0, 1, 3, 12, 60, 360, 2520, 20160, 181440, ... (Sloane's A001710).

출처 : http://mathworld.wolfram.com/OddPermutation.html

#### '★MATH > 2. 현대대수' 카테고리의 다른 글

 finite simple group  (0) 2010.04.07 2010.03.28 2010.03.25 2010.03.24 2010.03.22 2010.03.17

2010.03.24 14:48

## 군표현론 & homomorphism

2010.03.22 15:57

http://ko.wikipedia.org/wiki/%EA%B5%B0_%ED%91%9C%ED%98%84%EB%A1%A0
군표현론

http://en.wikipedia.org/wiki/Group_homomorphism
homomorphism

#### '★MATH > 2. 현대대수' 카테고리의 다른 글

 even & odd permutation  (0) 2010.03.25 2010.03.24 2010.03.22 2010.03.17 2010.03.17 2010.03.15

## cydlic group 2

2010.03.17 01:29

Cyclic Group Supplement
Theorem 1. Let g be an element of a group G and write
hgi = ngk : k 2 Zo.
Then hgi is a subgroup of G.
Proof. Since 1 = g0, 1 2 hgi. Suppose a, b 2 hgi. Then a = gk, b = gm and ab = gkgm = gk+m.
Hence ab 2 hgi (note that k + m 2 Z). Moreover, a−1 = (gk)−1 = g−k and −k 2 Z, so that
a−1 2 hgi. Thus, we have checked the three conditions necessary for hgi to be a subgroup of G.
Definition 2. If g 2 G, then the subgroup hgi = {gk : k 2 Z} is called the cyclic subgroup of G
generated by g, If G = hgi, then we say that G is a cyclic group and that g is a generator of
G.
Examples 3. 1. If G is any group then {1} = h1i is a cyclic subgroup of G.
2. The group G = {1, −1, i, −i} µ C¤ (the group operation is multiplication of complex numbers)
is cyclic with generator i. In fact hii = {i0 = 1, i1 = i, i2 = −1, i3 = −i} = G. Note that
−i is also a generator for G since h−ii = {(−i)0 = 1, (−i)1 = −i, (−i)2 = −1, (−i)3 = i} = G.
Thus a cyclic group may have more than one generator. However, not all elements of G need
be generators. For example h−1i = {1,−1} 6= G so −1 is not a generator of G.
3. The group G = Z¤
7 = the group of units of the ring Z7 is a cyclic group with generator 3.
Indeed,
h3i = {1 = 30, 3 = 31, 2 = 32, 6 = 33, 4 = 34, 5 = 35} = G.
Note that 5 is also a generator of G, but that h2i = {1, 2, 4} 6= G so that 2 is not a generator
of G.
4. G = h¼i = {¼k : k 2 Z} is a cyclic subgroup of R¤.
5. The group G = Z¤
8 is not cyclic. Indeed, since Z¤
8 = {1, 3, 5, 7} and h1i = {1}, h3i = {1, 3},
h5i = {1, 5}, h7i = {1, 7}, it follows that Z¤
8 6= hai for any a 2 Z¤
8.
If a group G is written additively, then the identity element is denoted 0, the inverse of a 2 G
is denoted −a, and the powers of a become na in additive notation. Thus, with this notation,
the cyclic subgroup of G generated by a is hai = {na : n 2 Z}, consisting of all the multiples of
a. Among groups that are normally written additively, the following are two examples of cyclic
groups.
6. The integers Z are a cyclic group. Indeed, Z = h1i since each integer k = k · 1 is a multiple
of 1, so k 2 h1i and h1i = Z. Also, Z = h−1i because k = (−k) · (−1) for each k 2 Z.
7. Zn is a cyclic group under addition with generator 1.
Theorem 4. Let g be an element of a group G. Then there are two possibilities for the cyclic
subgroup hgi.
Case 1: The cyclic subgroup hgi is finite. In this case, there exists a smallest positive integer
n such that gn = 1 and we have
(a) gk = 1 if and only if n|k.
(b) gk = gm if and only if k ´ m (mod n).
1
Cyclic Group Supplement
(c) hgi = {1, g, g2, . . . , gn−1} and the elements 1, g, g2, . . . , gn−1 are distinct.
Case 2: The cyclic subgroup hgi is infinite. Then
(d) gk = 1 if and only if k = 0.
(e) gk = gm if and only if k = m.
(f ) hgi = {. . . , g−3, g−2, g−1, 1, g, g2, g3, . . .} and all of these powers of g are distinct.
Proof. Case 1. Since hgi is finite, the powers g, g2, g3, . . . are not all distinct, so let gk = gm with
k < m. Then gm−k = 1 where m − k > 0. Hence there is a positive integer l with gl = 1. Hence
there is a smallest such positive integer. We let n be this smallest positive integer, i.e., n is the
smallest positive integer such that gn = 1.
(a) If n|k then k = qn for some q 2 n. Then gk = gqn = (gn)q = 1q = 1. Conversely, if gk = 1,
use the division algorithm to write k = qn + r with 0 · r < n. Then gr = gk(gn)−q = 1(1)−q = 1.
Since r < n, this contradicts the minimality of n unless r = 0. Hence r = 0 and k = qn so that n|k.
(b) gk = gm if and only if gk−m = 1. Now apply Part (a).
(c) Clearly, {1, g, g2, . . . , gn−1} µ hgi. To prove the other inclusion, let a 2 hgi. Then a = gk
for some k 2 Z. As in Part (a), use the division algorithm to write k = qn+r, where 0 · r · n−1.
Then
a = gk = gqn+r = (gn)qgr = 1qgr = gr 2 {1, g, g2, . . . , gn−1}
which shows that hgi µ {1, g, g2, . . . , gn−1}, and hence that
hgi = {1, g, g2, . . . , gn−1}.
Finally, suppose that gk = gm where 0 · k · m · n−1. Then gm−k = 1 and 0 · m−k < n. This
implies that m − k = 0 because n is the smallest positive power of g which equals 1. Hence all of
the elements 1, g, g2, . . . , gn−1 are distinct.
Case 2. (d) Certainly, gk = 1 if k = 0. If gk = 1, k 6= 0, then g−k = (gk)−1 = 1−1 = 1, also.
Hence gn = 1 for some n > 0, which implies that hgi is finite by the proof of Part (c), contrary to
our hypothesis in Case 2. Thus gk = 1 implies that k = 0.
(e) gk = gm if and only if gk−m = 1. Now apply Part (d).
(f) hgi = {gk : k 2 Z} by definition of hgi, so all that remains is to check that these powers are
distinct. But this is the content of Part (e).
Recall that if g is an element of a group G, then the order of g is the smallest positive integer
n such that gn = 1, and it is denoted o(g) = n. If there is no such positive integer, then we say
that g has infinite order, denoted o(g) = 1. By Theorem 4, the concept of order of an element
g and order of the cyclic subgroup generated by g are the same.
Corollary 5. If g is an element of a group G, then o(t) = |hgi|.
Proof. This is immediate from Theorem 4, Part (c).
If G is a cyclic group of order n, then it is easy to compute the order of all elements of G. This
is the content of the following result.
Theorem 6. Let G = hgi be a cyclic group of order n, and let 0 · k · n − 1. If m = gcd(k, n),
then o(gk) =
n
m
.
2
Cyclic Group Supplement
Proof. Let k = ms and n = mt. Then (gk)n/m = gkn/m = gmsn/m = (gn)s = 1s = 1. Hence n/m
divides o(gk) by Theorem 4 Part (a). Now suppose that (gk)r = 1. Then gkr = 1, so by Theorem
3 Part (a), n | kr. Hence
n
m¯¯¯
k
m
r
and since n/m and k/m are relatively prime, it follows that n/m divides r. Hence n/m is the
smallest power of gk which equals 1, so o(gk) = n/m.
Theorem 7. Let G = hgi be a cyclic group where o(g) = n. Then G = hgki if and only if
gcd(k, n) = 1.
Proof. By Theorem 6, if m = gcd(k, n), then o(gk) = n/m. But G = hgki if and only if o(gk) =
|G| = n and this happens if and only if m = 1, i.e., if and only if gcd(k, n) = 1.
Example 8. If G = hgi is a cyclic group of order 12, then the generators of G are the powers gk
where gcd(k, 12) = 1, that is g, g5, g7, and g11. In the particular case of the additive cyclic group
Z12, the generators are the integers 1, 5, 7, 11 (mod 12).
Now we ask what the subgroups of a cyclic group look like. The question is completely answered
by Theorem 10. Theorem 9 is a preliminary, but important, result.
Theorem 9. Every subgroup of a cyclic group is cyclic.
Proof. Suppose that G = hgi = {gk : k 2 Z} is a cyclic group and let H be a subgroup of G. If
H = {1}, then H is cyclic, so we assume that H 6= {1}, and let gk 2 H with gk 6= 1. Then, since H
is a subgroup, g−k = (gk)−1 2 H. Therefore, since k or −k is positive, H contains a positive power
of g, not equal to 1. So let m be the smallest positive integer such that gm 2 H. Then, certainly
all powers of gm are also in H, so we have hgmi µ H. We claim that this inclusion is an equality.
To see this, let gk be any element of H (recall that all elements of G, and hence H, are powers of
g since G is cyclic). By the division algorithm, we may write k = qm + r where 0 · r < m. But
gk = gqm+r = gqmgr = (gm)qgr so that
gr = (gm)−qgk 2 H.
Since m is the smallest positive integer with gm 2 H and 0 · r < m, it follows that we must have
r = 0. Then gk = (gm)q 2 hgmi. Hence we have shown that H µ hgmi and hence H = hgmi. That
is H is cyclic with generator gm where m is the smallest positive integer for which gm 2 H.
Theorem 10 (Fundamental Theorem of Finite Cyclic Groups). Let G = hgi be a cyclic
group of order n.
1. If H is any subgroup of G, then H = hgdi for some d|n.
2. If H is any subgroup of G with |H| = k, then k|n.
3. If k | n, then hgn/ki is the unique subgroup of G of order k.
Proof. 1. By Theorem 9, H is a cyclic group and since |G| = n < 1, it follows that H = hgmi
where m > 0. Let d = gcd(m, n). Since d | n it is sufficient to show that H = hgdi. But d|m
also, so m = qd. Then gm = (gd)q so gm 2 hgdi. Hence H = hgmi µ hgdi. But d = rm + sn,
where r, s 2 Z, so
gd = grm+sn = grmgsn = (gm)r(gn)s = (gm)r(1)s = (gm)r 2 hgmi = H.
This shows that hgdi µ H and hence hgdi = H.
3
Cyclic Group Supplement
2. By Part (a), H = hgdi where d | n. Then k = |H| = n/d so k|n.
3. Suppose that K is any subgroup of G of order k. By Part (a), let K = hgmi where m | n.
Then Theorem 6 gives k = |K| = |gm| = n/m. Hence m = n/k, so K = hgn/ki. This proves
(c).
Remark 11. Part (b) of Theorem 10 is actually true for any finite group G, whether or not it is
cyclic. This result is Lagrange’s Theorem (Theorem 5.2.3, Page 219 of your text).
The subgroups of a group G can be diagrammatically illustrated by listing the subgroups, and
indicating inclusion relations by means of a line directed upward from H to K if H is a subgroup
of K. Such a scheme is called the lattice diagram for the subgroups of the group G. We will
illustrate by determining the lattice diagram for all the subgroups of a cyclic group G = hgi of
order 12. Since the order of g is 12, Theorem 10 (c) shows that there is exactly one subgroup hgdi
for each divisor d of 12. The divisors of 12 are 1, 2, 3, 4, 6, 12. Then the unique subgroup of G of
each of these orders is, respectively,
{1} = hg12i, hg6i, hg4i, hg3i, hg2i, hgi = G.
Note that hgmi µ hgki if and only if k | m. Hence the lattice diagram of G is:
G
hg2i hg3i
hg4i hg6i
h1i
4

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## cyclic group

2010.03.17 00:46
```Date: 02/27/2003 at 17:11:41
From: Nicholas
Subject: Abstract Algebra/Group Theory

Let group G be finite Abelian such that G has the property that for
each positive integer n the set {x in G such that x^n = identity} has
at most n elements. Prove G is a cyclic group.
```

```Date: 03/01/2003 at 08:03:25
From: Doctor Jacques
Subject: Re: Abstract Algebra/Group Theory

Hi Nicholas,

In any finite group (say of n elements), every element x of the group
verifies x^n = e (this is a consequence of Lagrange's theorem).

A cyclic group is a group generated by a single element (for example
a). It is the set of powers of a:

{e, a, a^2, ......}

along with their inverses. We use the notation <a> for such a group.

If a cyclic group is finite, there is a smallest integer n such that
a^n = e, and the group consists of the elements:

<a> = {e, a, a^2, ..., a^(n-1)}

These n elements are all different, and the size of the group is
therefore n. As

a^i*a^(n-i) = a^n = e

we see that inverses are automatically included.

The order of an element is the order of the cyclic subgroup it
generates (in the above example, the order of a is n).

We also know that all elements x of <a> satisfy the equation x^n = e.

Sorry to repeat that if you already know it, I just wanted to be
sure.

Let us go back to the problem at hand.

What we are given is the fact that, for every n, the equation x^n = e
has at most n solutions.

As G is finite, we can find at least one element of maximal order. Let
a be such an element, <a> the subgroup it generates, and its order be
n.

If <a> = G, G is cyclic by definition and we are done.

Assume, for the sake of contradiction, that <a> is not equal to G.

We can therefore find an element b not in <a>. Let b be of order m.

As we chose n to be the maximal order of elements of G, we have
m <= n.

It is known that, in an Abelian group, if two elements have orders m
and n, there exists an element of order equal to:

p = lcm(m,n).

(If you are not familiar with that property, please write back, and
I'll explain it).

In this case, as n is the maximal order of an element, we must have
p = n (obviously, lcm(m,n) >= n).

The lcm of m and n will be equal to n if and only if m divides n.

This means that n = km for some k, and:

b^n = b^(km) = (b^m)^k = e^k = e.

and this shows that b is another solution of the equation x^n = e.

As all the n elements of <a> already satisfy that equation, and b is
not among them, the equation has at least n+1 solutions, contrary to
the hypothesis.

Does this help?  Write back if you'd like to talk about this
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```

```Date: 06/21/2003 at 03:00:15
From: William Cheung
Subject: Modern Algebra - Group Theory

Dear Dr. Jacques and associates, fellow mathematicians,

You wrote in a reply to a group theory problem:

It is known that, in an Abelian group, if two elements have orders m
and n, there exists an element of order equal to:

p = lcm(m,n).

I managed to solve the original problem, except that I couldn't prove
the omitted property related to the existence of the "lcm element."

I had the gut feeling that "lcm-order element" must exist in abelian
group, otherwise the original theorem would be wrong.  I can see that
there are two approaches to prove its existence:

1) constructive approach: for any two elements a, b in Abelian Group
G of order m and n, respectively, provide the closed form formula for
the "lcm-order element"

I tried for many weeks to arrive at such a formula, in vain. But my
best guess for the formula is, assume that gcd(m,n) = k, then gcd
(m/k, n/k) = 1, therefore there exist integer i and j such that
i * m + j * n = 1; the "lcm-order element" *may* probably be a^j * b^i
(usually multiplicative notation). I have a problem proving that it
is.

2) deductive approach: to conclude that it exists with some elegant
group-theoretic arguments.

Thank you for your kind reply.
```

```Date: 06/21/2003 at 08:14:48
From: Doctor Jacques
Subject: Re: Modern Algebra - Group Theory

Hi William,

Let us assume that a has order m and b has order n. We distinguish
two cases:

Case 1 : gcd(m,n) = 1
---------------------
In this case, lcm(m,n) = mn.

Consider the intersection H of <a> and <b>. H is a subgroup, and we
claim that H = {e}.

If x is an element of H, <x> is a subgroup of <a> and <b>, and so the
order of x must divide m and n, by Lagrange's theorem.

As gcd(m,n) = 1, x has order 1, i.e. x = e.

Let now x = a*b^(-1), and let k be the order of x.. As the group is
Abelian, x^(mn) = (a^m)*b^(-n) = e, and k divides mn.

We can write:

y^k = (a^k) * b^(-k) = e
a^k = b^(-k) = c

As c belongs to <a> and <b>, it belongs to H, and c = e. This means
that a^k = b^k = e, and k is a multiple of m and n, and therefore a
multiple of lcm(m,n) = mn. As we saw that k divides mn, this shows
that k = mn, and x is the required element.

Case 2 : gcd(m,n) = d > 1
-------------------------
We will factor m and n as m = m1*m2 and n = n1*n2.

Let p be a prime factor of mn (i.e. a prime factor of m or n), and
assume that p has exponent k1 in m and exponent k2 in n (one of these
exponents may be 0).

If k1 >= k2, include p^k1 in m1 and p^k2 in n2; otherwise, include
p^k1 in m2 and p^k2 in n1.

We can check the following:

* Every prime factor of mn appears in either m1 or n1, but not both,
which means that gcd(m1,n1) = 1.

* m1 and n1 contain all prime factors of mn with their greatest
exponent, and this shows that m1*n1 = lcm(m,n).

* Every prime power of m appears in either m1 or m2, but not both,
therefore m = m1*m2, and, in the same way, n = n1*n2.

For example, if m = 12 = 2^2*3 and n = 18 = 2*3^2, we will have

m1 = 2^2, m2 = 3
n1 = 3^2, n2 = 2

All this allows us to conclude that a^m2 and b^n2 have orders m1 and
n1, respectively, with gcd(m1,n1) = 1 and lcm(m1,n1) = lcm(m,n).

Can you continue from here? Write back if you'd like to talk about
this some more, or if you have any other questions.

```

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## 순환군

2010.03.15 21:54

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## 현대대수 솔루션

2010.03.06 12:41

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