Second isomorphism theorem on groups The Next CEO of Stack OverflowIntuition about the second isomorphism theoremThird isomorphism theorem on groupsFirst isomorphism theorem on groupsInterpretation of Second isomorphism theoremPlease check proof of Lagrange's theoremFind all groups such that there is a surjective homomorphismWhy $phi(H) cong H/ kerphi$ in the Second Isomorphism Theorem?Intuition about the second isomorphism theoremTwo normal subgroups and isomorphism theoremInner automorphisms and the permutation group $S_n$Find $dim(kerT_1 cap kerT_2)$ if $T_1,T_2:Vto F$, $T_1 neq 0$, $T_2neq 0$, $dimV=n$ and $kerT_1neq kerT_2$Let $G,H$ be groups and $varphi:G times Hto G$ and $H'=ker(varphi)$. Show that $(Gtimes H)/H'cong G$First isomorphism theorem on groups
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Second isomorphism theorem on groups
The Next CEO of Stack OverflowIntuition about the second isomorphism theoremThird isomorphism theorem on groupsFirst isomorphism theorem on groupsInterpretation of Second isomorphism theoremPlease check proof of Lagrange's theoremFind all groups such that there is a surjective homomorphismWhy $phi(H) cong H/ kerphi$ in the Second Isomorphism Theorem?Intuition about the second isomorphism theoremTwo normal subgroups and isomorphism theoremInner automorphisms and the permutation group $S_n$Find $dim(kerT_1 cap kerT_2)$ if $T_1,T_2:Vto F$, $T_1 neq 0$, $T_2neq 0$, $dimV=n$ and $kerT_1neq kerT_2$Let $G,H$ be groups and $varphi:G times Hto G$ and $H'=ker(varphi)$. Show that $(Gtimes H)/H'cong G$First isomorphism theorem on groups
$begingroup$
From this wikipedia's page, I have found these important theorems. I would like to give it a try. Since I am self-learning Mathematical Analysis without teacher or tutor, it will be great if someone helps me check it out. Thank you for your help!
Second isomorphism theorem on groups
Let $G$ be a group, $S$ a subgroup of $G$, and $N$ a normal subgroup of $G$.
$SN$ is a subgroup of $G$
$S cap N$ is a normal subgroup of $S$
$SN/N cong S/(S cap N)$
My attempt:
Let $s_1n_1, s_2n_2 in SN$. Since $N$ is normal, $n_1s_2 = s_2n'_1$ for some $n'_1 in N implies s_1n_1s_2 n_2 = s_1s_2n'_1n_2 in SN$. Let $sn in SN$. Since $N$ is normal, $sn = n's$ for some $n' in N implies (sn)^-1 = (n's)^-1 = s^-1(n')^-1 in SN$. Assertion (1.) then follows.
Clearly, $S cap N$ is a subgroup of $S$. Let $sin S$. If $h in s (S cap N)$, then $h = sm$ for some $m in S cap N$. Let $m' = sms^-1$. Then $m' in S$. Since $N$ is normal, $ms^-1 = s^-1 m''$ for some $m'' in N implies m' = ss^-1 m'' = m'' implies m' in S cap N implies (S cap N)s ni m' s = h implies$ $s(S cap N) subseteq (S cap N)s$. Similarly, $(S cap N)s subseteq s(S cap N)$. So $(S cap N)s = s(S cap N)$. Assertion (2.) then follows.
Clearly, $N$ is a normal subgroup of $SN$.
Approach 1:
Consider $$begin
arraylrcl
psi : & S/(S cap N)
& longrightarrow & SN/N\
& s(S cap N) & longmapsto & sN endarray$$
Clearly, $s(S cap N) subseteq sN$. If $s_1(S cap N) = s_2(S cap N)$ then $emptyset neq s_1(S cap N) subseteq (s_1N cap s_2N) implies s_1N = s_2N implies psi$ is well-defined. Let $snNin SN/N$. Then $snN=s(nN)=sN$ and thus $psi$ is surjective.
We have $psi (s_1(S cap N)) = psi (s_2(S cap N)) implies s_1N = s_2 N implies (s_1)^-1s_2 in N$ $implies (s_1)^-1s_2 in S cap N$. Moreover, $s_1((s_1)^-1s_2)=s_2e$ where $e in S cap N$ is the identity element of $G implies s_1(S cap N) cap s_2(S cap N) neq emptyset implies s_1(S cap N) =$ $s_2(S cap N) implies psi$ is injective.
We have $psi (s_1(S cap N) s_2(S cap N)) = psi (s_1s_2(S cap N)) = s_1s_2N = (s_1N) (s_2N) =$ $psi (s_1(S cap N)) psi (s_1(S cap N))$. Hence $psi$ is an isomorphism.
Approach 2:
Consider $$begin
arraylrcl
psi : & S
& longrightarrow & SN/N\
& s & longmapsto & sN endarray$$
Clearly, $sN=seN in SN/N$ for all $s in S$. Similarly, $psi$ is surjective. We have $psi (s_1 s_2) = (s_1 s_2)N = (s_1N)(s_2N) = psi(s_1) psi(s_2)$. Hence $psi$ is homorphism. We have $ker(psi) = s in S mid psi(s)= N = s in S mid sN = N$. Clearly, $s in N implies$ $s in ker(psi)$. If $s notin N$ then $sN cap N = emptyset$. Hence $ker(psi) = S cap N$.
By the first isomorphism theorem on groups, $S/ker(psi) = S/(S cap N) cong SN/N$.
group-theory proof-verification group-homomorphism
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
$endgroup$
add a comment |
$begingroup$
From this wikipedia's page, I have found these important theorems. I would like to give it a try. Since I am self-learning Mathematical Analysis without teacher or tutor, it will be great if someone helps me check it out. Thank you for your help!
Second isomorphism theorem on groups
Let $G$ be a group, $S$ a subgroup of $G$, and $N$ a normal subgroup of $G$.
$SN$ is a subgroup of $G$
$S cap N$ is a normal subgroup of $S$
$SN/N cong S/(S cap N)$
My attempt:
Let $s_1n_1, s_2n_2 in SN$. Since $N$ is normal, $n_1s_2 = s_2n'_1$ for some $n'_1 in N implies s_1n_1s_2 n_2 = s_1s_2n'_1n_2 in SN$. Let $sn in SN$. Since $N$ is normal, $sn = n's$ for some $n' in N implies (sn)^-1 = (n's)^-1 = s^-1(n')^-1 in SN$. Assertion (1.) then follows.
Clearly, $S cap N$ is a subgroup of $S$. Let $sin S$. If $h in s (S cap N)$, then $h = sm$ for some $m in S cap N$. Let $m' = sms^-1$. Then $m' in S$. Since $N$ is normal, $ms^-1 = s^-1 m''$ for some $m'' in N implies m' = ss^-1 m'' = m'' implies m' in S cap N implies (S cap N)s ni m' s = h implies$ $s(S cap N) subseteq (S cap N)s$. Similarly, $(S cap N)s subseteq s(S cap N)$. So $(S cap N)s = s(S cap N)$. Assertion (2.) then follows.
Clearly, $N$ is a normal subgroup of $SN$.
Approach 1:
Consider $$begin
arraylrcl
psi : & S/(S cap N)
& longrightarrow & SN/N\
& s(S cap N) & longmapsto & sN endarray$$
Clearly, $s(S cap N) subseteq sN$. If $s_1(S cap N) = s_2(S cap N)$ then $emptyset neq s_1(S cap N) subseteq (s_1N cap s_2N) implies s_1N = s_2N implies psi$ is well-defined. Let $snNin SN/N$. Then $snN=s(nN)=sN$ and thus $psi$ is surjective.
We have $psi (s_1(S cap N)) = psi (s_2(S cap N)) implies s_1N = s_2 N implies (s_1)^-1s_2 in N$ $implies (s_1)^-1s_2 in S cap N$. Moreover, $s_1((s_1)^-1s_2)=s_2e$ where $e in S cap N$ is the identity element of $G implies s_1(S cap N) cap s_2(S cap N) neq emptyset implies s_1(S cap N) =$ $s_2(S cap N) implies psi$ is injective.
We have $psi (s_1(S cap N) s_2(S cap N)) = psi (s_1s_2(S cap N)) = s_1s_2N = (s_1N) (s_2N) =$ $psi (s_1(S cap N)) psi (s_1(S cap N))$. Hence $psi$ is an isomorphism.
Approach 2:
Consider $$begin
arraylrcl
psi : & S
& longrightarrow & SN/N\
& s & longmapsto & sN endarray$$
Clearly, $sN=seN in SN/N$ for all $s in S$. Similarly, $psi$ is surjective. We have $psi (s_1 s_2) = (s_1 s_2)N = (s_1N)(s_2N) = psi(s_1) psi(s_2)$. Hence $psi$ is homorphism. We have $ker(psi) = s in S mid psi(s)= N = s in S mid sN = N$. Clearly, $s in N implies$ $s in ker(psi)$. If $s notin N$ then $sN cap N = emptyset$. Hence $ker(psi) = S cap N$.
By the first isomorphism theorem on groups, $S/ker(psi) = S/(S cap N) cong SN/N$.
group-theory proof-verification group-homomorphism
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
$endgroup$
$begingroup$
Also compare, like for the first one, with the standard proofs at this site, e.g. here.
$endgroup$
– Dietrich Burde
Mar 20 at 10:28
add a comment |
$begingroup$
From this wikipedia's page, I have found these important theorems. I would like to give it a try. Since I am self-learning Mathematical Analysis without teacher or tutor, it will be great if someone helps me check it out. Thank you for your help!
Second isomorphism theorem on groups
Let $G$ be a group, $S$ a subgroup of $G$, and $N$ a normal subgroup of $G$.
$SN$ is a subgroup of $G$
$S cap N$ is a normal subgroup of $S$
$SN/N cong S/(S cap N)$
My attempt:
Let $s_1n_1, s_2n_2 in SN$. Since $N$ is normal, $n_1s_2 = s_2n'_1$ for some $n'_1 in N implies s_1n_1s_2 n_2 = s_1s_2n'_1n_2 in SN$. Let $sn in SN$. Since $N$ is normal, $sn = n's$ for some $n' in N implies (sn)^-1 = (n's)^-1 = s^-1(n')^-1 in SN$. Assertion (1.) then follows.
Clearly, $S cap N$ is a subgroup of $S$. Let $sin S$. If $h in s (S cap N)$, then $h = sm$ for some $m in S cap N$. Let $m' = sms^-1$. Then $m' in S$. Since $N$ is normal, $ms^-1 = s^-1 m''$ for some $m'' in N implies m' = ss^-1 m'' = m'' implies m' in S cap N implies (S cap N)s ni m' s = h implies$ $s(S cap N) subseteq (S cap N)s$. Similarly, $(S cap N)s subseteq s(S cap N)$. So $(S cap N)s = s(S cap N)$. Assertion (2.) then follows.
Clearly, $N$ is a normal subgroup of $SN$.
Approach 1:
Consider $$begin
arraylrcl
psi : & S/(S cap N)
& longrightarrow & SN/N\
& s(S cap N) & longmapsto & sN endarray$$
Clearly, $s(S cap N) subseteq sN$. If $s_1(S cap N) = s_2(S cap N)$ then $emptyset neq s_1(S cap N) subseteq (s_1N cap s_2N) implies s_1N = s_2N implies psi$ is well-defined. Let $snNin SN/N$. Then $snN=s(nN)=sN$ and thus $psi$ is surjective.
We have $psi (s_1(S cap N)) = psi (s_2(S cap N)) implies s_1N = s_2 N implies (s_1)^-1s_2 in N$ $implies (s_1)^-1s_2 in S cap N$. Moreover, $s_1((s_1)^-1s_2)=s_2e$ where $e in S cap N$ is the identity element of $G implies s_1(S cap N) cap s_2(S cap N) neq emptyset implies s_1(S cap N) =$ $s_2(S cap N) implies psi$ is injective.
We have $psi (s_1(S cap N) s_2(S cap N)) = psi (s_1s_2(S cap N)) = s_1s_2N = (s_1N) (s_2N) =$ $psi (s_1(S cap N)) psi (s_1(S cap N))$. Hence $psi$ is an isomorphism.
Approach 2:
Consider $$begin
arraylrcl
psi : & S
& longrightarrow & SN/N\
& s & longmapsto & sN endarray$$
Clearly, $sN=seN in SN/N$ for all $s in S$. Similarly, $psi$ is surjective. We have $psi (s_1 s_2) = (s_1 s_2)N = (s_1N)(s_2N) = psi(s_1) psi(s_2)$. Hence $psi$ is homorphism. We have $ker(psi) = s in S mid psi(s)= N = s in S mid sN = N$. Clearly, $s in N implies$ $s in ker(psi)$. If $s notin N$ then $sN cap N = emptyset$. Hence $ker(psi) = S cap N$.
By the first isomorphism theorem on groups, $S/ker(psi) = S/(S cap N) cong SN/N$.
group-theory proof-verification group-homomorphism
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
$endgroup$
From this wikipedia's page, I have found these important theorems. I would like to give it a try. Since I am self-learning Mathematical Analysis without teacher or tutor, it will be great if someone helps me check it out. Thank you for your help!
Second isomorphism theorem on groups
Let $G$ be a group, $S$ a subgroup of $G$, and $N$ a normal subgroup of $G$.
$SN$ is a subgroup of $G$
$S cap N$ is a normal subgroup of $S$
$SN/N cong S/(S cap N)$
My attempt:
Let $s_1n_1, s_2n_2 in SN$. Since $N$ is normal, $n_1s_2 = s_2n'_1$ for some $n'_1 in N implies s_1n_1s_2 n_2 = s_1s_2n'_1n_2 in SN$. Let $sn in SN$. Since $N$ is normal, $sn = n's$ for some $n' in N implies (sn)^-1 = (n's)^-1 = s^-1(n')^-1 in SN$. Assertion (1.) then follows.
Clearly, $S cap N$ is a subgroup of $S$. Let $sin S$. If $h in s (S cap N)$, then $h = sm$ for some $m in S cap N$. Let $m' = sms^-1$. Then $m' in S$. Since $N$ is normal, $ms^-1 = s^-1 m''$ for some $m'' in N implies m' = ss^-1 m'' = m'' implies m' in S cap N implies (S cap N)s ni m' s = h implies$ $s(S cap N) subseteq (S cap N)s$. Similarly, $(S cap N)s subseteq s(S cap N)$. So $(S cap N)s = s(S cap N)$. Assertion (2.) then follows.
Clearly, $N$ is a normal subgroup of $SN$.
Approach 1:
Consider $$begin
arraylrcl
psi : & S/(S cap N)
& longrightarrow & SN/N\
& s(S cap N) & longmapsto & sN endarray$$
Clearly, $s(S cap N) subseteq sN$. If $s_1(S cap N) = s_2(S cap N)$ then $emptyset neq s_1(S cap N) subseteq (s_1N cap s_2N) implies s_1N = s_2N implies psi$ is well-defined. Let $snNin SN/N$. Then $snN=s(nN)=sN$ and thus $psi$ is surjective.
We have $psi (s_1(S cap N)) = psi (s_2(S cap N)) implies s_1N = s_2 N implies (s_1)^-1s_2 in N$ $implies (s_1)^-1s_2 in S cap N$. Moreover, $s_1((s_1)^-1s_2)=s_2e$ where $e in S cap N$ is the identity element of $G implies s_1(S cap N) cap s_2(S cap N) neq emptyset implies s_1(S cap N) =$ $s_2(S cap N) implies psi$ is injective.
We have $psi (s_1(S cap N) s_2(S cap N)) = psi (s_1s_2(S cap N)) = s_1s_2N = (s_1N) (s_2N) =$ $psi (s_1(S cap N)) psi (s_1(S cap N))$. Hence $psi$ is an isomorphism.
Approach 2:
Consider $$begin
arraylrcl
psi : & S
& longrightarrow & SN/N\
& s & longmapsto & sN endarray$$
Clearly, $sN=seN in SN/N$ for all $s in S$. Similarly, $psi$ is surjective. We have $psi (s_1 s_2) = (s_1 s_2)N = (s_1N)(s_2N) = psi(s_1) psi(s_2)$. Hence $psi$ is homorphism. We have $ker(psi) = s in S mid psi(s)= N = s in S mid sN = N$. Clearly, $s in N implies$ $s in ker(psi)$. If $s notin N$ then $sN cap N = emptyset$. Hence $ker(psi) = S cap N$.
By the first isomorphism theorem on groups, $S/ker(psi) = S/(S cap N) cong SN/N$.
group-theory proof-verification group-homomorphism
group-theory proof-verification group-homomorphism
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
asked Mar 20 at 10:10
Le Anh DungLe Anh Dung
1,4461621
1,4461621
$begingroup$
Also compare, like for the first one, with the standard proofs at this site, e.g. here.
$endgroup$
– Dietrich Burde
Mar 20 at 10:28
add a comment |
$begingroup$
Also compare, like for the first one, with the standard proofs at this site, e.g. here.
$endgroup$
– Dietrich Burde
Mar 20 at 10:28
$begingroup$
Also compare, like for the first one, with the standard proofs at this site, e.g. here.
$endgroup$
– Dietrich Burde
Mar 20 at 10:28
$begingroup$
Also compare, like for the first one, with the standard proofs at this site, e.g. here.
$endgroup$
– Dietrich Burde
Mar 20 at 10:28
add a comment |
0
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$begingroup$
Also compare, like for the first one, with the standard proofs at this site, e.g. here.
$endgroup$
– Dietrich Burde
Mar 20 at 10:28