Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ; forall g in G$, $g= h_{1}h_{2}ldots h_n$ where...
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Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
> forall g in G$, $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
unique. Here, $H_{i}lhd G ; forall i$
I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.
group-theory direct-product
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Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
> forall g in G$, $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
unique. Here, $H_{i}lhd G ; forall i$
I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.
group-theory direct-product
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
> forall g in G$, $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
unique. Here, $H_{i}lhd G ; forall i$
I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.
group-theory direct-product
Let $G=H_{1} times H_{2} dots times H_{n}$ if and only if $ ;
> forall g in G$, $g= h_{1}h_{2}ldots h_n$ where $h_{i} in H_{i}$ is
unique. Here, $H_{i}lhd G ; forall i$
I have proved this for the case when we have only $2$ normal groups, $(n=2)$ but I am finding it difficult to prove the general result. I feel like I am missing some important intermediate results.
group-theory direct-product
group-theory direct-product
asked Nov 21 at 4:51
So Lo
62218
62218
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Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.
Take base case for n=2 . which you had already proved
Assume this result for n such normal group direct product
then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.
Take base case for n=2 . which you had already proved
Assume this result for n such normal group direct product
then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now
add a comment |
up vote
1
down vote
Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.
Take base case for n=2 . which you had already proved
Assume this result for n such normal group direct product
then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now
add a comment |
up vote
1
down vote
up vote
1
down vote
Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.
Take base case for n=2 . which you had already proved
Assume this result for n such normal group direct product
then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now
Hint: As you told you have done this for 2 normal subgroup then you can tackle other by mathematical induction.
Take base case for n=2 . which you had already proved
Assume this result for n such normal group direct product
then to prove for n+1 take assumed normal group as single piece normal subgroup and this problem reduces to 2 now
answered Nov 21 at 5:15
Shubham
1,3711518
1,3711518
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