Decompose $G$ into product of simple groups











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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?






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  • Yes, it is correct.
    – freakish
    yesterday















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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







share|cite|improve this question






















  • Yes, it is correct.
    – freakish
    yesterday













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







share|cite|improve this question













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






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  • Yes, it is correct.
    – freakish
    yesterday


















  • Yes, it is correct.
    – freakish
    yesterday
















Yes, it is correct.
– freakish
yesterday




Yes, it is correct.
– freakish
yesterday















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