g*****e
发帖数: 574 | 1 show that a group of order 2m is not simple, where m is an odd integer. | W**********d
发帖数: 15 | 2 Well, I guess Z_2 is a simple group. Suppose your group has a subgroup with
order m, it is not simple.
【在 g*****e 的大作中提到】
: show that a group of order 2m is not simple, where m is an odd integer.
| v*******e
发帖数: 3714 | 3 这么一说我才想起，
我连simple的定义都已经忘干净了。
【在 g*****e 的大作中提到】
: show that a group of order 2m is not simple, where m is an odd integer.
| a*******o
发帖数: 2 | 4 Let G be your group with order 2m, where m is odd.
It is well known that any even-ordered finite group contains
an element of order 2. Let H be the set of elements x in G
satisfying x^2 = e (the identity element). Then the order of
H is an even number and a divisor of 2m. Thus, |H| = 2 or |H| = 2m.
If |H| = 2m, then H = G. One can then show that G is abelian
and, with order =2m, not simple.
If |H| = 2, then one can show that H is a nontrivial
normal subgroup of G. Thus G is not simple.
Some deta | b****d
发帖数: 1311 | 5 Have you checked the example S_3?
Moreover, someone gave a counterexample Z_2 recently.
The proof of Feit-Thompson's theorem: Every odd order finite group is
solvable, is not trivial. For order 2m with m>1, even if the statement
is true, it may be hard to prove.
【在 a*******o 的大作中提到】
: Let G be your group with order 2m, where m is odd.
: It is well known that any even-ordered finite group contains
: an element of order 2. Let H be the set of elements x in G
: satisfying x^2 = e (the identity element). Then the order of
: H is an even number and a divisor of 2m. Thus, |H| = 2 or |H| = 2m.
: If |H| = 2m, then H = G. One can then show that G is abelian
: and, with order =2m, not simple.
: If |H| = 2, then one can show that H is a nontrivial
: normal subgroup of G. Thus G is not simple.
: Some deta
| g******a
发帖数: 69 | 6 by considering the pair (a,a^{-1}),
we can find g_1\in G such that g_1\neq e and g_1^2=1.
Next, let A be the commutation group of G, i.e. |A|=(2m)!.
For any g\in G, map the action on G of left multiplication
by g to A: g\to T_g\in A, T_g(h)=gh.
T(g_1) is a commutation of odd order.
Let G_1 be a subgroup of G such that T(g) is of even order for
any g\in G_1. It is easy to check that G_1 is a normal group
and |G_1|=m, since if T(g) is of even order, then T(gg_1) is
of odd order.
【在 b****d 的大作中提到】
: Have you checked the example S_3?
: Moreover, someone gave a counterexample Z_2 recently.
: The proof of Feit-Thompson's theorem: Every odd order finite group is
: solvable, is not trivial. For order 2m with m>1, even if the statement
: is true, it may be hard to prove.
| y**t
发帖数: 50 | 7 I think you mean permutation group not commutation.
And also the problem needs to add a condition that m > 1.
It's just that the order 2 element is an odd permutation (product of m 2-cycles)
in G's regular representation (left or right multiplication). And then
the elements in G which are even permutations in the representation
consists of a subgroup in G with index 2.
It's an easy execise that index 2 subgroup is a normal group.
【在 g******a 的大作中提到】
: by considering the pair (a,a^{-1}),
: we can find g_1\in G such that g_1\neq e and g_1^2=1.
: Next, let A be the commutation group of G, i.e. |A|=(2m)!.
: For any g\in G, map the action on G of left multiplication
: by g to A: g\to T_g\in A, T_g(h)=gh.
: T(g_1) is a commutation of odd order.
: Let G_1 be a subgroup of G such that T(g) is of even order for
: any g\in G_1. It is easy to check that G_1 is a normal group
: and |G_1|=m, since if T(g) is of even order, then T(gg_1) is
: of odd order.
| g******a
发帖数: 69 | 8 yes, permutation.
cycles)
【在 y**t 的大作中提到】
: I think you mean permutation group not commutation.
: And also the problem needs to add a condition that m > 1.
: It's just that the order 2 element is an odd permutation (product of m 2-cycles)
: in G's regular representation (left or right multiplication). And then
: the elements in G which are even permutations in the representation
: consists of a subgroup in G with index 2.
: It's an easy execise that index 2 subgroup is a normal group.
| b****d
发帖数: 1311 | 9
Thanks. A nice proof!
【在 g******a 的大作中提到】
: by considering the pair (a,a^{-1}),
: we can find g_1\in G such that g_1\neq e and g_1^2=1.
: Next, let A be the commutation group of G, i.e. |A|=(2m)!.
: For any g\in G, map the action on G of left multiplication
: by g to A: g\to T_g\in A, T_g(h)=gh.
: T(g_1) is a commutation of odd order.
: Let G_1 be a subgroup of G such that T(g) is of even order for
: any g\in G_1. It is easy to check that G_1 is a normal group
: and |G_1|=m, since if T(g) is of even order, then T(gg_1) is
: of odd order.
| a*******o
发帖数: 2 | 10 It is well known that S_3 is not a simple group.
As a matter of fact, the smallest nonabelian simple group is the alternating
group A_5 of order 60. |
|
