Third isomorphism theorem on groupsSecond isomorphism theorem on groupsAre normal subgroups transitive?Interpretation of First Isomorphism TheoremWhy $phi(H) cong H/ kerphi$ in the Second Isomorphism Theorem?Question in line of proof for first isomorphism theoremIntuition about the first isomorphism theoremIntuition about the second isomorphism theoremShow normal subgroup $N$ is a subset of $kerphi$ if $psi$ is well-definedLet $G,H$ be groups and $varphi:G times Hto G$ and $H'=ker(varphi)$. Show that $(Gtimes H)/H'cong G$First isomorphism theorem on groupsSecond isomorphism theorem on groups
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Third isomorphism theorem on groups
Second isomorphism theorem on groupsAre normal subgroups transitive?Interpretation of First Isomorphism TheoremWhy $phi(H) cong H/ kerphi$ in the Second Isomorphism Theorem?Question in line of proof for first isomorphism theoremIntuition about the first isomorphism theoremIntuition about the second isomorphism theoremShow normal subgroup $N$ is a subset of $kerphi$ if $psi$ is well-definedLet $G,H$ be groups and $varphi:G times Hto G$ and $H'=ker(varphi)$. Show that $(Gtimes H)/H'cong G$First isomorphism theorem on groupsSecond 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!
Third isomorphism theorem on groups
Let $G$ be a group and $N$ a normal subgroup of $G$.
If $K$ is a subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a subgroup of $G/N$.
Every subgroup of $G/N$ is of the form $K/N$, for some subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a normal subgroup of $G/N$.
Every normal subgroup of $G/N$ is of the form $K/N$, for some normal subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then the quotient group $(G/N)/(K/N) cong G/K$.
My attempt:
Clearly, $K/N,G/N$ are groups and $K/N subseteq G/N$. So $K/N$ is a subgroup of $G/N$. Assertion (1.) then follows.
Let $H$ be a subgroup of $G/N$ and $K = g in G mid gN in H$. If $g_1, g_2 in K$ then $g_1N in H,g_2N in H$. So $(g_1N)(g_2N) in H$ or $(g_1g_2)N in H$. Hence $g_1g_2 in K$ and thus $K$ is closed. If $g in K$ then $gN in H$ and thus $g^-1N = Ng^-1 = N^-1g^-1 = (gN)^-1 in H$. Hence $g^-1 in K$. So $K$ is a group. Clearly, $H = K/N$. Since $N$ is the identity element of $G/N$ and $H$ a subgroup of $G/N$, $N in H$. We have $n in N implies nN=N in H implies$ $n in K implies N subseteq K$. It follows that $H = K/N$. Assertion (2.) then follows.
Because of (1.), it suffices to prove that $K/N$ is normal in $G/N$. For all $kN in K/N,gN in G/N$, $(gN)(kN)(gN)^-1 = gNNkN^-1g^-1 = gNkNg^-1 =$ $gNNkg^-1 = gNkg^-1 = gkNg^-1 =$ $gkg^-1N in K/N.$ Assertion (3.) then follows.
Because of (2.), it suffices to prove that $K$ is normal in $G$. Since $K/N$ is normal in $G/N$, $forall (kN in K/N, gN in G/N): (kN)(gN)(kN)^-1 in K/N$. On the other hand, $(kN)(gN)(kN)^-1 in K/N iff kNNgN^-1k^-1 in K/N iff kNgNk^-1 in K/N$ $iff$ $kgNNk^-1 in K/N iff kgNk^-1 in K/N iff kgk^-1N in K/N iff$ $kgk^-1 in K$. This implies $forall (k in K, gin G): kgk^-1 in K$ and thus $K$ is normal in G. Assertion (4.) then follows.
Consider
$$begin
arraylrcl
psi : & G/N
& longrightarrow & G/K\
& gN & longmapsto & gK endarray$$
Clearly, $psi$ is surjective. We have $$psi((g_1N)(g_2N)) = psi(g_1g_2N) = g_1g_2K = (g_1K)(g_2K) =psi(g_1N) psi(g_2N)$$ Hence $psi$ is a homomorphism.
$ker (psi) = gN in G/N mid psi(gN) = K = gN in G/N mid gK = K$. On the other hand, $gK=K iff g in K$. Hence $ker (psi) =K/N$.
By the first isomorphism theorem on groups, we have $$(G/N)/ker(psi) = (G/N)/(K/N) cong G/K$$ Assertion (5.) then follows.
group-theory proof-verification group-homomorphism
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
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!
Third isomorphism theorem on groups
Let $G$ be a group and $N$ a normal subgroup of $G$.
If $K$ is a subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a subgroup of $G/N$.
Every subgroup of $G/N$ is of the form $K/N$, for some subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a normal subgroup of $G/N$.
Every normal subgroup of $G/N$ is of the form $K/N$, for some normal subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then the quotient group $(G/N)/(K/N) cong G/K$.
My attempt:
Clearly, $K/N,G/N$ are groups and $K/N subseteq G/N$. So $K/N$ is a subgroup of $G/N$. Assertion (1.) then follows.
Let $H$ be a subgroup of $G/N$ and $K = g in G mid gN in H$. If $g_1, g_2 in K$ then $g_1N in H,g_2N in H$. So $(g_1N)(g_2N) in H$ or $(g_1g_2)N in H$. Hence $g_1g_2 in K$ and thus $K$ is closed. If $g in K$ then $gN in H$ and thus $g^-1N = Ng^-1 = N^-1g^-1 = (gN)^-1 in H$. Hence $g^-1 in K$. So $K$ is a group. Clearly, $H = K/N$. Since $N$ is the identity element of $G/N$ and $H$ a subgroup of $G/N$, $N in H$. We have $n in N implies nN=N in H implies$ $n in K implies N subseteq K$. It follows that $H = K/N$. Assertion (2.) then follows.
Because of (1.), it suffices to prove that $K/N$ is normal in $G/N$. For all $kN in K/N,gN in G/N$, $(gN)(kN)(gN)^-1 = gNNkN^-1g^-1 = gNkNg^-1 =$ $gNNkg^-1 = gNkg^-1 = gkNg^-1 =$ $gkg^-1N in K/N.$ Assertion (3.) then follows.
Because of (2.), it suffices to prove that $K$ is normal in $G$. Since $K/N$ is normal in $G/N$, $forall (kN in K/N, gN in G/N): (kN)(gN)(kN)^-1 in K/N$. On the other hand, $(kN)(gN)(kN)^-1 in K/N iff kNNgN^-1k^-1 in K/N iff kNgNk^-1 in K/N$ $iff$ $kgNNk^-1 in K/N iff kgNk^-1 in K/N iff kgk^-1N in K/N iff$ $kgk^-1 in K$. This implies $forall (k in K, gin G): kgk^-1 in K$ and thus $K$ is normal in G. Assertion (4.) then follows.
Consider
$$begin
arraylrcl
psi : & G/N
& longrightarrow & G/K\
& gN & longmapsto & gK endarray$$
Clearly, $psi$ is surjective. We have $$psi((g_1N)(g_2N)) = psi(g_1g_2N) = g_1g_2K = (g_1K)(g_2K) =psi(g_1N) psi(g_2N)$$ Hence $psi$ is a homomorphism.
$ker (psi) = gN in G/N mid psi(gN) = K = gN in G/N mid gK = K$. On the other hand, $gK=K iff g in K$. Hence $ker (psi) =K/N$.
By the first isomorphism theorem on groups, we have $$(G/N)/ker(psi) = (G/N)/(K/N) cong G/K$$ Assertion (5.) then follows.
group-theory proof-verification group-homomorphism
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
$begingroup$
See also first and second isomorphism theorem here.
$endgroup$
– Dietrich Burde
Mar 21 at 12:18
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!
Third isomorphism theorem on groups
Let $G$ be a group and $N$ a normal subgroup of $G$.
If $K$ is a subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a subgroup of $G/N$.
Every subgroup of $G/N$ is of the form $K/N$, for some subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a normal subgroup of $G/N$.
Every normal subgroup of $G/N$ is of the form $K/N$, for some normal subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then the quotient group $(G/N)/(K/N) cong G/K$.
My attempt:
Clearly, $K/N,G/N$ are groups and $K/N subseteq G/N$. So $K/N$ is a subgroup of $G/N$. Assertion (1.) then follows.
Let $H$ be a subgroup of $G/N$ and $K = g in G mid gN in H$. If $g_1, g_2 in K$ then $g_1N in H,g_2N in H$. So $(g_1N)(g_2N) in H$ or $(g_1g_2)N in H$. Hence $g_1g_2 in K$ and thus $K$ is closed. If $g in K$ then $gN in H$ and thus $g^-1N = Ng^-1 = N^-1g^-1 = (gN)^-1 in H$. Hence $g^-1 in K$. So $K$ is a group. Clearly, $H = K/N$. Since $N$ is the identity element of $G/N$ and $H$ a subgroup of $G/N$, $N in H$. We have $n in N implies nN=N in H implies$ $n in K implies N subseteq K$. It follows that $H = K/N$. Assertion (2.) then follows.
Because of (1.), it suffices to prove that $K/N$ is normal in $G/N$. For all $kN in K/N,gN in G/N$, $(gN)(kN)(gN)^-1 = gNNkN^-1g^-1 = gNkNg^-1 =$ $gNNkg^-1 = gNkg^-1 = gkNg^-1 =$ $gkg^-1N in K/N.$ Assertion (3.) then follows.
Because of (2.), it suffices to prove that $K$ is normal in $G$. Since $K/N$ is normal in $G/N$, $forall (kN in K/N, gN in G/N): (kN)(gN)(kN)^-1 in K/N$. On the other hand, $(kN)(gN)(kN)^-1 in K/N iff kNNgN^-1k^-1 in K/N iff kNgNk^-1 in K/N$ $iff$ $kgNNk^-1 in K/N iff kgNk^-1 in K/N iff kgk^-1N in K/N iff$ $kgk^-1 in K$. This implies $forall (k in K, gin G): kgk^-1 in K$ and thus $K$ is normal in G. Assertion (4.) then follows.
Consider
$$begin
arraylrcl
psi : & G/N
& longrightarrow & G/K\
& gN & longmapsto & gK endarray$$
Clearly, $psi$ is surjective. We have $$psi((g_1N)(g_2N)) = psi(g_1g_2N) = g_1g_2K = (g_1K)(g_2K) =psi(g_1N) psi(g_2N)$$ Hence $psi$ is a homomorphism.
$ker (psi) = gN in G/N mid psi(gN) = K = gN in G/N mid gK = K$. On the other hand, $gK=K iff g in K$. Hence $ker (psi) =K/N$.
By the first isomorphism theorem on groups, we have $$(G/N)/ker(psi) = (G/N)/(K/N) cong G/K$$ Assertion (5.) then follows.
group-theory proof-verification group-homomorphism
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
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!
Third isomorphism theorem on groups
Let $G$ be a group and $N$ a normal subgroup of $G$.
If $K$ is a subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a subgroup of $G/N$.
Every subgroup of $G/N$ is of the form $K/N$, for some subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then $K/N$ is a normal subgroup of $G/N$.
Every normal subgroup of $G/N$ is of the form $K/N$, for some normal subgroup $K$ of $G$ such that $N subseteq K subseteq G$.
If $K$ is a normal subgroup of $G$ such that $N subseteq K subseteq G$, then the quotient group $(G/N)/(K/N) cong G/K$.
My attempt:
Clearly, $K/N,G/N$ are groups and $K/N subseteq G/N$. So $K/N$ is a subgroup of $G/N$. Assertion (1.) then follows.
Let $H$ be a subgroup of $G/N$ and $K = g in G mid gN in H$. If $g_1, g_2 in K$ then $g_1N in H,g_2N in H$. So $(g_1N)(g_2N) in H$ or $(g_1g_2)N in H$. Hence $g_1g_2 in K$ and thus $K$ is closed. If $g in K$ then $gN in H$ and thus $g^-1N = Ng^-1 = N^-1g^-1 = (gN)^-1 in H$. Hence $g^-1 in K$. So $K$ is a group. Clearly, $H = K/N$. Since $N$ is the identity element of $G/N$ and $H$ a subgroup of $G/N$, $N in H$. We have $n in N implies nN=N in H implies$ $n in K implies N subseteq K$. It follows that $H = K/N$. Assertion (2.) then follows.
Because of (1.), it suffices to prove that $K/N$ is normal in $G/N$. For all $kN in K/N,gN in G/N$, $(gN)(kN)(gN)^-1 = gNNkN^-1g^-1 = gNkNg^-1 =$ $gNNkg^-1 = gNkg^-1 = gkNg^-1 =$ $gkg^-1N in K/N.$ Assertion (3.) then follows.
Because of (2.), it suffices to prove that $K$ is normal in $G$. Since $K/N$ is normal in $G/N$, $forall (kN in K/N, gN in G/N): (kN)(gN)(kN)^-1 in K/N$. On the other hand, $(kN)(gN)(kN)^-1 in K/N iff kNNgN^-1k^-1 in K/N iff kNgNk^-1 in K/N$ $iff$ $kgNNk^-1 in K/N iff kgNk^-1 in K/N iff kgk^-1N in K/N iff$ $kgk^-1 in K$. This implies $forall (k in K, gin G): kgk^-1 in K$ and thus $K$ is normal in G. Assertion (4.) then follows.
Consider
$$begin
arraylrcl
psi : & G/N
& longrightarrow & G/K\
& gN & longmapsto & gK endarray$$
Clearly, $psi$ is surjective. We have $$psi((g_1N)(g_2N)) = psi(g_1g_2N) = g_1g_2K = (g_1K)(g_2K) =psi(g_1N) psi(g_2N)$$ Hence $psi$ is a homomorphism.
$ker (psi) = gN in G/N mid psi(gN) = K = gN in G/N mid gK = K$. On the other hand, $gK=K iff g in K$. Hence $ker (psi) =K/N$.
By the first isomorphism theorem on groups, we have $$(G/N)/ker(psi) = (G/N)/(K/N) cong G/K$$ Assertion (5.) then follows.
group-theory proof-verification group-homomorphism
group-theory proof-verification group-homomorphism
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
edited Mar 21 at 5:59
Le Anh Dung
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
asked Mar 21 at 5:07
Le Anh DungLe Anh Dung
1,4511621
1,4511621
$begingroup$
See also first and second isomorphism theorem here.
$endgroup$
– Dietrich Burde
Mar 21 at 12:18
add a comment |
$begingroup$
See also first and second isomorphism theorem here.
$endgroup$
– Dietrich Burde
Mar 21 at 12:18
$begingroup$
See also first and second isomorphism theorem here.
$endgroup$
– Dietrich Burde
Mar 21 at 12:18
$begingroup$
See also first and second isomorphism theorem here.
$endgroup$
– Dietrich Burde
Mar 21 at 12:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your proof(s) seem fine to me.
However, you have used "clearly" three times. Whenever you feel the urge to use such a word in mathematics, imagine a complete beginner asking, "why?" - because these are the places we most often hide gaps in our own understanding. Don't fool yourself!
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
$endgroup$
add a comment |
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1 Answer
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$begingroup$
Your proof(s) seem fine to me.
However, you have used "clearly" three times. Whenever you feel the urge to use such a word in mathematics, imagine a complete beginner asking, "why?" - because these are the places we most often hide gaps in our own understanding. Don't fool yourself!
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
$endgroup$
add a comment |
$begingroup$
Your proof(s) seem fine to me.
However, you have used "clearly" three times. Whenever you feel the urge to use such a word in mathematics, imagine a complete beginner asking, "why?" - because these are the places we most often hide gaps in our own understanding. Don't fool yourself!
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
$endgroup$
add a comment |
$begingroup$
Your proof(s) seem fine to me.
However, you have used "clearly" three times. Whenever you feel the urge to use such a word in mathematics, imagine a complete beginner asking, "why?" - because these are the places we most often hide gaps in our own understanding. Don't fool yourself!
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
$endgroup$
Your proof(s) seem fine to me.
However, you have used "clearly" three times. Whenever you feel the urge to use such a word in mathematics, imagine a complete beginner asking, "why?" - because these are the places we most often hide gaps in our own understanding. Don't fool yourself!
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
edited Mar 21 at 15:10
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
answered Mar 21 at 10:26
ShaunShaun
10.1k113685
10.1k113685
add a comment |
add a comment |
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$begingroup$
See also first and second isomorphism theorem here.
$endgroup$
– Dietrich Burde
Mar 21 at 12:18