Mixing
${prod_{i=1}^{r} H_{i}} bigcap H_{r+1} = {e} Rightarrow H_{i}bigcap H_{j} = {e} ; forall ineq j$?
I am studying Internal Direct Products in Group Theory.
$mbox{Let}$ $G$ be a group and $H_{i}lhd G; forall i=1,2ldots
> n$
Suppose ${prod_{i=1}^{r} H_{i}} bigcap H_{r+1} = {e} ; forall
> r=1,2ldots (n-1)$ then prove that $H_{i}bigcap H_{j} = {e} ;
forall ineq j$
But converse is true only if $n=2$
First of all, I am really not sure about the validity of these statements since I am reading from a not-so-standard material.
I tried to prove $(Rightarrow)$ like this$:$
Let $ain H_{i}bigcap H_{j},; ; aneq e, ; ; f.s. ; ineq j$
$Rightarrow ain H_{i}$ and $a in H_{j}$
WLOG, assume $i < j $
Clearly, $a in prod_{i=1}^{j-1}H_{i}$ since we can write $a=e.eldots a.e.dots e$
$Rightarrow {prod_{i=1}^{j-1}H_{i} } bigcap H_{j} neq {e} ; ; mbox{contradiction}$
Converse is obviously true for $n=2$ but how can I show it is not true for $n>2$?
group-theory direct-product
asked Nov 21 '18 at 4:30
So Lo
63719
add a comment |
I am studying Internal Direct Products in Group Theory.
$mbox{Let}$ $G$ be a group and $H_{i}lhd G; forall i=1,2ldots
> n$
Suppose ${prod_{i=1}^{r} H_{i}} bigcap H_{r+1} = {e} ; forall
> r=1,2ldots (n-1)$ then prove that $H_{i}bigcap H_{j} = {e} ;
forall ineq j$
But converse is true only if $n=2$
First of all, I am really not sure about the validity of these statements since I am reading from a not-so-standard material.
I tried to prove $(Rightarrow)$ like this$:$
Let $ain H_{i}bigcap H_{j},; ; aneq e, ; ; f.s. ; ineq j$
$Rightarrow ain H_{i}$ and $a in H_{j}$
WLOG, assume $i < j $
Clearly, $a in prod_{i=1}^{j-1}H_{i}$ since we can write $a=e.eldots a.e.dots e$
$Rightarrow {prod_{i=1}^{j-1}H_{i} } bigcap H_{j} neq {e} ; ; mbox{contradiction}$
Converse is obviously true for $n=2$ but how can I show it is not true for $n>2$?
group-theory direct-product
asked Nov 21 '18 at 4:30
So Lo
63719
Have you tried to prove this theorem... solo? :)
– Monstrous Moonshiner
Nov 21 '18 at 6:08
@MonstrousMoonshiner I've proved one side of it in the description itself
– So Lo
Nov 21 '18 at 7:10
Yes, I was just making a silly joke...
– Monstrous Moonshiner
Nov 21 '18 at 7:11
@MonstrousMoonshiner ohh, my bad :p
– So Lo
Nov 21 '18 at 7:11
add a comment |
I am studying Internal Direct Products in Group Theory.
$mbox{Let}$ $G$ be a group and $H_{i}lhd G; forall i=1,2ldots
> n$
Suppose ${prod_{i=1}^{r} H_{i}} bigcap H_{r+1} = {e} ; forall
> r=1,2ldots (n-1)$ then prove that $H_{i}bigcap H_{j} = {e} ;
forall ineq j$
But converse is true only if $n=2$
First of all, I am really not sure about the validity of these statements since I am reading from a not-so-standard material.
I tried to prove $(Rightarrow)$ like this$:$
Let $ain H_{i}bigcap H_{j},; ; aneq e, ; ; f.s. ; ineq j$
$Rightarrow ain H_{i}$ and $a in H_{j}$
WLOG, assume $i < j $
Clearly, $a in prod_{i=1}^{j-1}H_{i}$ since we can write $a=e.eldots a.e.dots e$
$Rightarrow {prod_{i=1}^{j-1}H_{i} } bigcap H_{j} neq {e} ; ; mbox{contradiction}$
Converse is obviously true for $n=2$ but how can I show it is not true for $n>2$?
group-theory direct-product
asked Nov 21 '18 at 4:30
So Lo
63719
I am studying Internal Direct Products in Group Theory.
$mbox{Let}$ $G$ be a group and $H_{i}lhd G; forall i=1,2ldots
> n$
Suppose ${prod_{i=1}^{r} H_{i}} bigcap H_{r+1} = {e} ; forall
> r=1,2ldots (n-1)$ then prove that $H_{i}bigcap H_{j} = {e} ;
forall ineq j$
But converse is true only if $n=2$
First of all, I am really not sure about the validity of these statements since I am reading from a not-so-standard material.
I tried to prove $(Rightarrow)$ like this$:$
Let $ain H_{i}bigcap H_{j},; ; aneq e, ; ; f.s. ; ineq j$
$Rightarrow ain H_{i}$ and $a in H_{j}$
WLOG, assume $i < j $
Clearly, $a in prod_{i=1}^{j-1}H_{i}$ since we can write $a=e.eldots a.e.dots e$
$Rightarrow {prod_{i=1}^{j-1}H_{i} } bigcap H_{j} neq {e} ; ; mbox{contradiction}$
Converse is obviously true for $n=2$ but how can I show it is not true for $n>2$?
group-theory direct-product
group-theory direct-product
asked Nov 21 '18 at 4:30
So Lo
63719
asked Nov 21 '18 at 4:30
So Lo
63719
asked Nov 21 '18 at 4:30
So Lo
63719
asked Nov 21 '18 at 4:30
So Lo
63719
asked Nov 21 '18 at 4:30
So Lo
63719
63719
Have you tried to prove this theorem... solo? :)
– Monstrous Moonshiner
Nov 21 '18 at 6:08
@MonstrousMoonshiner I've proved one side of it in the description itself
– So Lo
Nov 21 '18 at 7:10
Yes, I was just making a silly joke...
– Monstrous Moonshiner
Nov 21 '18 at 7:11
@MonstrousMoonshiner ohh, my bad :p
– So Lo
Nov 21 '18 at 7:11
add a comment |
Have you tried to prove this theorem... solo? :)
– Monstrous Moonshiner
Nov 21 '18 at 6:08
@MonstrousMoonshiner I've proved one side of it in the description itself
– So Lo
Nov 21 '18 at 7:10
Yes, I was just making a silly joke...
– Monstrous Moonshiner
Nov 21 '18 at 7:11
@MonstrousMoonshiner ohh, my bad :p
– So Lo
Nov 21 '18 at 7:11
Have you tried to prove this theorem... solo? :)
– Monstrous Moonshiner
Nov 21 '18 at 6:08
Have you tried to prove this theorem... solo? :)
– Monstrous Moonshiner
Nov 21 '18 at 6:08
@MonstrousMoonshiner I've proved one side of it in the description itself
– So Lo
Nov 21 '18 at 7:10
@MonstrousMoonshiner I've proved one side of it in the description itself
– So Lo
Nov 21 '18 at 7:10
Yes, I was just making a silly joke...
– Monstrous Moonshiner
Nov 21 '18 at 7:11
Yes, I was just making a silly joke...
– Monstrous Moonshiner
Nov 21 '18 at 7:11
@MonstrousMoonshiner ohh, my bad :p
– So Lo
Nov 21 '18 at 7:11
@MonstrousMoonshiner ohh, my bad :p
– So Lo
Nov 21 '18 at 7:11
add a comment |
1 Answer
1
active
oldest
votes
Counter example :
Group $Z_3times Z_3$ under +
$N_1=${$(0,0),(1,0),(2,0)$}
$N_2=${$(0,0)(0,1)(0,2)$}
$N_3=${$(0,0)(1,1)(2,2)$}
$N_1N_2cap N_3=${$(0,0),(1,1)(2,2)$}
answered Nov 21 '18 at 5:28
Shubham
1,5951519
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Counter example :
Group $Z_3times Z_3$ under +
$N_1=${$(0,0),(1,0),(2,0)$}
$N_2=${$(0,0)(0,1)(0,2)$}
$N_3=${$(0,0)(1,1)(2,2)$}
$N_1N_2cap N_3=${$(0,0),(1,1)(2,2)$}
answered Nov 21 '18 at 5:28
Shubham
1,5951519
add a comment |
Counter example :
Group $Z_3times Z_3$ under +
$N_1=${$(0,0),(1,0),(2,0)$}
$N_2=${$(0,0)(0,1)(0,2)$}
$N_3=${$(0,0)(1,1)(2,2)$}
$N_1N_2cap N_3=${$(0,0),(1,1)(2,2)$}
answered Nov 21 '18 at 5:28
Shubham
1,5951519
add a comment |
Counter example :
Group $Z_3times Z_3$ under +
$N_1=${$(0,0),(1,0),(2,0)$}
$N_2=${$(0,0)(0,1)(0,2)$}
$N_3=${$(0,0)(1,1)(2,2)$}
$N_1N_2cap N_3=${$(0,0),(1,1)(2,2)$}
answered Nov 21 '18 at 5:28
Shubham
1,5951519
Counter example :
Group $Z_3times Z_3$ under +
$N_1=${$(0,0),(1,0),(2,0)$}
$N_2=${$(0,0)(0,1)(0,2)$}
$N_3=${$(0,0)(1,1)(2,2)$}
$N_1N_2cap N_3=${$(0,0),(1,1)(2,2)$}
answered Nov 21 '18 at 5:28
Shubham
1,5951519
edited Nov 21 '18 at 5:56
answered Nov 21 '18 at 5:28
Shubham
1,5951519
answered Nov 21 '18 at 5:28
Shubham
1,5951519
answered Nov 21 '18 at 5:28
Shubham
1,5951519
1,5951519
add a comment |
add a comment |
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Have you tried to prove this theorem... solo? :)
– Monstrous Moonshiner
Nov 21 '18 at 6:08
@MonstrousMoonshiner I've proved one side of it in the description itself
– So Lo
Nov 21 '18 at 7:10
Yes, I was just making a silly joke...
– Monstrous Moonshiner
Nov 21 '18 at 7:11
@MonstrousMoonshiner ohh, my bad :p
– So Lo
Nov 21 '18 at 7:11