Show that $(Mtimes N)/Rcong M$.
$begingroup$
If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that
$eR = text{identity element}$
$R = {(eR, r) | r in R}$
Show that
$$
frac{M times N}{R} simeq M
$$
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 24 '18 at 11:21
Shaun
9,356113684
asked Dec 24 '18 at 4:20
user628956user628956
203
$endgroup$
add a comment |
$begingroup$
If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that
$eR = text{identity element}$
$R = {(eR, r) | r in R}$
Show that
$$
frac{M times N}{R} simeq M
$$
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 24 '18 at 11:21
Shaun
9,356113684
asked Dec 24 '18 at 4:20
user628956user628956
203
$endgroup$
$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24
$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25
$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28
$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30
add a comment |
$begingroup$
If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that
$eR = text{identity element}$
$R = {(eR, r) | r in R}$
Show that
$$
frac{M times N}{R} simeq M
$$
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 24 '18 at 11:21
Shaun
9,356113684
asked Dec 24 '18 at 4:20
user628956user628956
203
$endgroup$
If there are two groups, $M$ (with multiplication $cdot_M$) and $N$ (with multiplication $cdot_N$) and we define a new group $M times N$ with multiplication such that
$$
(m,n)(m',n') = (m cdot_M m',n cdot_N n')
$$
There can be a normal subgroup $R$ of $M times N$ such that
$eR = text{identity element}$
$R = {(eR, r) | r in R}$
Show that
$$
frac{M times N}{R} simeq M
$$
abstract-algebra group-theory normal-subgroups group-isomorphism
abstract-algebra group-theory normal-subgroups group-isomorphism
edited Dec 24 '18 at 11:21
Shaun
9,356113684
asked Dec 24 '18 at 4:20
user628956user628956
203
edited Dec 24 '18 at 11:21
Shaun
9,356113684
asked Dec 24 '18 at 4:20
user628956user628956
203
edited Dec 24 '18 at 11:21
Shaun
9,356113684
edited Dec 24 '18 at 11:21
Shaun
9,356113684
edited Dec 24 '18 at 11:21
Shaun
9,356113684
9,356113684
asked Dec 24 '18 at 4:20
user628956user628956
203
asked Dec 24 '18 at 4:20
user628956user628956
203
asked Dec 24 '18 at 4:20
user628956user628956
203
203
$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24
$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25
$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28
$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30
add a comment |
$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24
$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25
$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28
$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30
$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24
$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24
$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25
$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25
$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28
$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28
$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30
$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:
$$
(M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
$$
You can check it to be a group homomorphism.
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
$endgroup$
add a comment |
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1 Answer
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oldest
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1 Answer
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active
oldest
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votes
$begingroup$
Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:
$$
(M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
$$
You can check it to be a group homomorphism.
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
$endgroup$
add a comment |
$begingroup$
Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:
$$
(M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
$$
You can check it to be a group homomorphism.
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
$endgroup$
add a comment |
$begingroup$
Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:
$$
(M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
$$
You can check it to be a group homomorphism.
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
$endgroup$
Use the first isomorphism theorem under the mapping $phi: M times N to M$ that takes $(m,n) mapsto m$. This map is surjective onto $M$, and has kernel $R$ (since whenever $n = e$, we get mapped to the identity). Thus:
$$
(M times N) / ker(phi) cong Im(phi) implies (M times N)/R cong M
$$
You can check it to be a group homomorphism.
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
answered Dec 24 '18 at 4:26
TrostAftTrostAft
423412
423412
add a comment |
add a comment |
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$begingroup$
Doesn't first isomorphism theorem do the trick under the mapping $(m, n) mapsto m$?
$endgroup$
– TrostAft
Dec 24 '18 at 4:24
$begingroup$
What about R - the normal subgroup?
$endgroup$
– user628956
Dec 24 '18 at 4:25
$begingroup$
Right so FIT states if we have a group homomorphism $phi$ from $M to N$, then: $M / ker(phi) cong Im(N)$. That gives us exactly what we want, where $ker(phi)$ gives us $R$.
$endgroup$
– TrostAft
Dec 24 '18 at 4:28
$begingroup$
Also I hastily ended up answering this, but please provide what you tried next time.
$endgroup$
– TrostAft
Dec 24 '18 at 4:30