Mixing
Decompose $G$ into product of simple groups
up vote
0
down vote
favorite
prove that If $G=H_1times...times H_m=K_1times....times K_n$, where each $H_i$ and $K_i$ are simple groups then $m=n$ and there is a permutation $sigma in S_n$ such that $H_i cong K_{sigma(i)}$ for all $i=1,2,...,n.$
Here is my attempt:
each $H_i$ is normal in $G$, $G$ has composition series: $G=G_0=H_1times...times H_m triangleright G_1=H_2 times ...times H_m triangleright ...triangleright G_{m-1}=H_m triangleright G_m={1}$.
also:
$G=M_0=K_1times...times K_n triangleright M_1=K_2 times ...times K_n triangleright ...triangleright M_{n-1}=K_n triangleright M_n={1}$.
So by Jordan Holder theorem, $n=m$, and the composition factors of the two composition series are permutations of each others. Also, $H_i$ and $K_i$ can be made into composition factors in a composition serie that ends with $H_i triangleright {1}$ or $K_i triangleright {1}$. so that means we can construct two composition series so that $H_i cong K_{sigma(i)}$. Is this proof correct?
group-theory
asked yesterday
mathnoob
62811
add a comment |
up vote
0
down vote
favorite
prove that If $G=H_1times...times H_m=K_1times....times K_n$, where each $H_i$ and $K_i$ are simple groups then $m=n$ and there is a permutation $sigma in S_n$ such that $H_i cong K_{sigma(i)}$ for all $i=1,2,...,n.$
Here is my attempt:
each $H_i$ is normal in $G$, $G$ has composition series: $G=G_0=H_1times...times H_m triangleright G_1=H_2 times ...times H_m triangleright ...triangleright G_{m-1}=H_m triangleright G_m={1}$.
also:
$G=M_0=K_1times...times K_n triangleright M_1=K_2 times ...times K_n triangleright ...triangleright M_{n-1}=K_n triangleright M_n={1}$.
So by Jordan Holder theorem, $n=m$, and the composition factors of the two composition series are permutations of each others. Also, $H_i$ and $K_i$ can be made into composition factors in a composition serie that ends with $H_i triangleright {1}$ or $K_i triangleright {1}$. so that means we can construct two composition series so that $H_i cong K_{sigma(i)}$. Is this proof correct?
group-theory
asked yesterday
mathnoob
62811
Yes, it is correct.
– freakish
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
prove that If $G=H_1times...times H_m=K_1times....times K_n$, where each $H_i$ and $K_i$ are simple groups then $m=n$ and there is a permutation $sigma in S_n$ such that $H_i cong K_{sigma(i)}$ for all $i=1,2,...,n.$
Here is my attempt:
each $H_i$ is normal in $G$, $G$ has composition series: $G=G_0=H_1times...times H_m triangleright G_1=H_2 times ...times H_m triangleright ...triangleright G_{m-1}=H_m triangleright G_m={1}$.
also:
$G=M_0=K_1times...times K_n triangleright M_1=K_2 times ...times K_n triangleright ...triangleright M_{n-1}=K_n triangleright M_n={1}$.
So by Jordan Holder theorem, $n=m$, and the composition factors of the two composition series are permutations of each others. Also, $H_i$ and $K_i$ can be made into composition factors in a composition serie that ends with $H_i triangleright {1}$ or $K_i triangleright {1}$. so that means we can construct two composition series so that $H_i cong K_{sigma(i)}$. Is this proof correct?
group-theory
asked yesterday
mathnoob
62811
prove that If $G=H_1times...times H_m=K_1times....times K_n$, where each $H_i$ and $K_i$ are simple groups then $m=n$ and there is a permutation $sigma in S_n$ such that $H_i cong K_{sigma(i)}$ for all $i=1,2,...,n.$
Here is my attempt:
each $H_i$ is normal in $G$, $G$ has composition series: $G=G_0=H_1times...times H_m triangleright G_1=H_2 times ...times H_m triangleright ...triangleright G_{m-1}=H_m triangleright G_m={1}$.
also:
$G=M_0=K_1times...times K_n triangleright M_1=K_2 times ...times K_n triangleright ...triangleright M_{n-1}=K_n triangleright M_n={1}$.
So by Jordan Holder theorem, $n=m$, and the composition factors of the two composition series are permutations of each others. Also, $H_i$ and $K_i$ can be made into composition factors in a composition serie that ends with $H_i triangleright {1}$ or $K_i triangleright {1}$. so that means we can construct two composition series so that $H_i cong K_{sigma(i)}$. Is this proof correct?
group-theory
group-theory
asked yesterday
mathnoob
62811
asked yesterday
mathnoob
62811
asked yesterday
mathnoob
62811
asked yesterday
mathnoob
62811
asked yesterday
mathnoob
62811
62811
Yes, it is correct.
– freakish
yesterday
add a comment |
Yes, it is correct.
– freakish
yesterday
Yes, it is correct.
– freakish
yesterday
Yes, it is correct.
– freakish
yesterday
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005199%2fdecompose-g-into-product-of-simple-groups%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Yes, it is correct.
– freakish
yesterday