Mixing
Let $N_1, . . . , N_n$ be normal subgroups of $G$, consider $G/N_1times ··· times G/N_n$
$begingroup$
I’m reading Hans Kurzweil ‘s “The Theory of Finite Groups”, where it says
1.6.4 Let $N_1, . . . , N_n$ be normal subgroups of $G$. Then the mapping $$α: G→G/N_1times ··· times G/N_n$$ given by $$g mapsto
(gN_1,...,gN_n)$$ is a homomorphism with $operatorname{Ker}α = cap_i N_i$. In
particular, $G/cap_i N_i$ is isomorphic to a subgroup of $G/N_1
times ··· times G/N_n$.
I’m confused here: can we write $$G/N_1times cdots times G/N_n$$
? To write a product of groups as this, it’s required that each $G/N_i$ has only $e$ as common element.
What if $$G=C_2 times C_3 times C_5 times C_7$$
$$N_1=C_2 times C_3 $$
$$N_2=C_2 times C_5 $$
$$N_3=C_2 times C_7 $$
, shouldn’t $$G/N_1 cong C_3 times C_5$$
$$G/N_2 cong C_2 times C_7$$
$$G/N_3 cong C_5 times C_7$$
, and they have common elements besides $e$?
abstract-algebra group-theory
edited Jan 8 at 11:39
user1729
16.9k64085
asked Jan 8 at 10:38
athosathos
86511339
$endgroup$
add a comment |
$begingroup$
I’m reading Hans Kurzweil ‘s “The Theory of Finite Groups”, where it says
1.6.4 Let $N_1, . . . , N_n$ be normal subgroups of $G$. Then the mapping $$α: G→G/N_1times ··· times G/N_n$$ given by $$g mapsto
(gN_1,...,gN_n)$$ is a homomorphism with $operatorname{Ker}α = cap_i N_i$. In
particular, $G/cap_i N_i$ is isomorphic to a subgroup of $G/N_1
times ··· times G/N_n$.
I’m confused here: can we write $$G/N_1times cdots times G/N_n$$
? To write a product of groups as this, it’s required that each $G/N_i$ has only $e$ as common element.
What if $$G=C_2 times C_3 times C_5 times C_7$$
$$N_1=C_2 times C_3 $$
$$N_2=C_2 times C_5 $$
$$N_3=C_2 times C_7 $$
, shouldn’t $$G/N_1 cong C_3 times C_5$$
$$G/N_2 cong C_2 times C_7$$
$$G/N_3 cong C_5 times C_7$$
, and they have common elements besides $e$?
abstract-algebra group-theory
edited Jan 8 at 11:39
user1729
16.9k64085
asked Jan 8 at 10:38
athosathos
86511339
$endgroup$
$begingroup$
I believe this is the Chinese Reminder theorem for groups.
$endgroup$
– IAmNoOne
Jan 9 at 13:21
$begingroup$
@IAmNoOne but there’s no solution formula?
$endgroup$
– athos
Jan 9 at 23:55
add a comment |
$begingroup$
I’m reading Hans Kurzweil ‘s “The Theory of Finite Groups”, where it says
1.6.4 Let $N_1, . . . , N_n$ be normal subgroups of $G$. Then the mapping $$α: G→G/N_1times ··· times G/N_n$$ given by $$g mapsto
(gN_1,...,gN_n)$$ is a homomorphism with $operatorname{Ker}α = cap_i N_i$. In
particular, $G/cap_i N_i$ is isomorphic to a subgroup of $G/N_1
times ··· times G/N_n$.
I’m confused here: can we write $$G/N_1times cdots times G/N_n$$
? To write a product of groups as this, it’s required that each $G/N_i$ has only $e$ as common element.
What if $$G=C_2 times C_3 times C_5 times C_7$$
$$N_1=C_2 times C_3 $$
$$N_2=C_2 times C_5 $$
$$N_3=C_2 times C_7 $$
, shouldn’t $$G/N_1 cong C_3 times C_5$$
$$G/N_2 cong C_2 times C_7$$
$$G/N_3 cong C_5 times C_7$$
, and they have common elements besides $e$?
abstract-algebra group-theory
edited Jan 8 at 11:39
user1729
16.9k64085
asked Jan 8 at 10:38
athosathos
86511339
$endgroup$
I’m reading Hans Kurzweil ‘s “The Theory of Finite Groups”, where it says
1.6.4 Let $N_1, . . . , N_n$ be normal subgroups of $G$. Then the mapping $$α: G→G/N_1times ··· times G/N_n$$ given by $$g mapsto
(gN_1,...,gN_n)$$ is a homomorphism with $operatorname{Ker}α = cap_i N_i$. In
particular, $G/cap_i N_i$ is isomorphic to a subgroup of $G/N_1
times ··· times G/N_n$.
I’m confused here: can we write $$G/N_1times cdots times G/N_n$$
? To write a product of groups as this, it’s required that each $G/N_i$ has only $e$ as common element.
What if $$G=C_2 times C_3 times C_5 times C_7$$
$$N_1=C_2 times C_3 $$
$$N_2=C_2 times C_5 $$
$$N_3=C_2 times C_7 $$
, shouldn’t $$G/N_1 cong C_3 times C_5$$
$$G/N_2 cong C_2 times C_7$$
$$G/N_3 cong C_5 times C_7$$
, and they have common elements besides $e$?
abstract-algebra group-theory
abstract-algebra group-theory
edited Jan 8 at 11:39
user1729
16.9k64085
asked Jan 8 at 10:38
athosathos
86511339
edited Jan 8 at 11:39
user1729
16.9k64085
asked Jan 8 at 10:38
athosathos
86511339
edited Jan 8 at 11:39
user1729
16.9k64085
edited Jan 8 at 11:39
user1729
16.9k64085
edited Jan 8 at 11:39
user1729
16.9k64085
16.9k64085
asked Jan 8 at 10:38
athosathos
86511339
asked Jan 8 at 10:38
athosathos
86511339
asked Jan 8 at 10:38
athosathos
86511339
86511339
$begingroup$
I believe this is the Chinese Reminder theorem for groups.
$endgroup$
– IAmNoOne
Jan 9 at 13:21
$begingroup$
@IAmNoOne but there’s no solution formula?
$endgroup$
– athos
Jan 9 at 23:55
add a comment |
$begingroup$
I believe this is the Chinese Reminder theorem for groups.
$endgroup$
– IAmNoOne
Jan 9 at 13:21
$begingroup$
@IAmNoOne but there’s no solution formula?
$endgroup$
– athos
Jan 9 at 23:55
$begingroup$
I believe this is the Chinese Reminder theorem for groups.
$endgroup$
– IAmNoOne
Jan 9 at 13:21
$begingroup$
I believe this is the Chinese Reminder theorem for groups.
$endgroup$
– IAmNoOne
Jan 9 at 13:21
$begingroup$
@IAmNoOne but there’s no solution formula?
$endgroup$
– athos
Jan 9 at 23:55
$begingroup$
@IAmNoOne but there’s no solution formula?
$endgroup$
– athos
Jan 9 at 23:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I’m confused here: can we write $$G/N_1times ··· times G/N_n$$ ? To
write a product of groups as this, it’s required that each $G/N_i$ has
only $e$ as common element.
Note that $G/N_i$ is not a subgroup of $G $. So here we are not considering the internal direct product, which requires the condition you mentioned above to be a group. Here $G/N_1times ··· times G/N_n$ represents the external direct product , which is a group under the componentwise operation.
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
$endgroup$
add a comment |
$begingroup$
Given two groups $G_1$ and $G_2$, you can form their direct product: $G_1×G_2$, to be ${(g_1,g_2)mid g_1in G_1,g_2in G_2}$, with the group operation defined as $(g_1,g_2)+(h_1,h_2)=(g_1+h_1,g_2+h_2)$.
As an example, consider the Klein four group, $V_4=C_2×C_2$.
(As in the other answer, you seem to be thinking of the internal direct product of subgroups of the given group.)
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
$endgroup$
add a comment |
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2 Answers
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2 Answers
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votes
$begingroup$
I’m confused here: can we write $$G/N_1times ··· times G/N_n$$ ? To
write a product of groups as this, it’s required that each $G/N_i$ has
only $e$ as common element.
Note that $G/N_i$ is not a subgroup of $G $. So here we are not considering the internal direct product, which requires the condition you mentioned above to be a group. Here $G/N_1times ··· times G/N_n$ represents the external direct product , which is a group under the componentwise operation.
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
$endgroup$
add a comment |
$begingroup$
I’m confused here: can we write $$G/N_1times ··· times G/N_n$$ ? To
write a product of groups as this, it’s required that each $G/N_i$ has
only $e$ as common element.
Note that $G/N_i$ is not a subgroup of $G $. So here we are not considering the internal direct product, which requires the condition you mentioned above to be a group. Here $G/N_1times ··· times G/N_n$ represents the external direct product , which is a group under the componentwise operation.
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
$endgroup$
add a comment |
$begingroup$
I’m confused here: can we write $$G/N_1times ··· times G/N_n$$ ? To
write a product of groups as this, it’s required that each $G/N_i$ has
only $e$ as common element.
Note that $G/N_i$ is not a subgroup of $G $. So here we are not considering the internal direct product, which requires the condition you mentioned above to be a group. Here $G/N_1times ··· times G/N_n$ represents the external direct product , which is a group under the componentwise operation.
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
$endgroup$
I’m confused here: can we write $$G/N_1times ··· times G/N_n$$ ? To
write a product of groups as this, it’s required that each $G/N_i$ has
only $e$ as common element.
Note that $G/N_i$ is not a subgroup of $G $. So here we are not considering the internal direct product, which requires the condition you mentioned above to be a group. Here $G/N_1times ··· times G/N_n$ represents the external direct product , which is a group under the componentwise operation.
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
answered Jan 8 at 11:10
Thomas ShelbyThomas Shelby
2,700421
2,700421
add a comment |
add a comment |
$begingroup$
Given two groups $G_1$ and $G_2$, you can form their direct product: $G_1×G_2$, to be ${(g_1,g_2)mid g_1in G_1,g_2in G_2}$, with the group operation defined as $(g_1,g_2)+(h_1,h_2)=(g_1+h_1,g_2+h_2)$.
As an example, consider the Klein four group, $V_4=C_2×C_2$.
(As in the other answer, you seem to be thinking of the internal direct product of subgroups of the given group.)
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
$endgroup$
add a comment |
$begingroup$
Given two groups $G_1$ and $G_2$, you can form their direct product: $G_1×G_2$, to be ${(g_1,g_2)mid g_1in G_1,g_2in G_2}$, with the group operation defined as $(g_1,g_2)+(h_1,h_2)=(g_1+h_1,g_2+h_2)$.
As an example, consider the Klein four group, $V_4=C_2×C_2$.
(As in the other answer, you seem to be thinking of the internal direct product of subgroups of the given group.)
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
$endgroup$
add a comment |
$begingroup$
Given two groups $G_1$ and $G_2$, you can form their direct product: $G_1×G_2$, to be ${(g_1,g_2)mid g_1in G_1,g_2in G_2}$, with the group operation defined as $(g_1,g_2)+(h_1,h_2)=(g_1+h_1,g_2+h_2)$.
As an example, consider the Klein four group, $V_4=C_2×C_2$.
(As in the other answer, you seem to be thinking of the internal direct product of subgroups of the given group.)
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
$endgroup$
Given two groups $G_1$ and $G_2$, you can form their direct product: $G_1×G_2$, to be ${(g_1,g_2)mid g_1in G_1,g_2in G_2}$, with the group operation defined as $(g_1,g_2)+(h_1,h_2)=(g_1+h_1,g_2+h_2)$.
As an example, consider the Klein four group, $V_4=C_2×C_2$.
(As in the other answer, you seem to be thinking of the internal direct product of subgroups of the given group.)
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
answered Jan 8 at 11:35
Chris CusterChris Custer
12k3825
12k3825
add a comment |
add a comment |
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$begingroup$
I believe this is the Chinese Reminder theorem for groups.
$endgroup$
– IAmNoOne
Jan 9 at 13:21
$begingroup$
@IAmNoOne but there’s no solution formula?
$endgroup$
– athos
Jan 9 at 23:55