Prove that if $h_{1},h_{2} in H$ and $k_{1},k_{2} in K$ and $h_{1}k_{1}=h_{2}k_{2}$ , then $h_{1}=h_{2}$ and...
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Prove that if $h_{1},h_{2} in H$ and $k_{1},k_{2} in K$ and $h_{1}k_{1}=h_{2}k_{2}$ , then $h_{1}=h_{2}$ and $k_{1}=k_{2}$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Proof that $T_i unlhd T$Subgroups and IndexProve that $G$ is finite and $|G|$ is prime.Prove that the groups $H/(Hcap K)$ and $HK/K$ are isomorphic.Suppose that $x,y,z$ are elements of a group and are conjugate. Prove for $H,N vartriangleleft G$ and $x in H$ and $y in N$ that $zin Hcap N$.If the intersection of two normal subgroups is trivial, then their elements commuteProve that the group of order 3 is cyclic.Proof that $G$ is the semi direct product of $P$ and $Q$ if and only if the composition $phicirc iota : Prightarrow G/Q$ is an isomorphismProve that $K={e}$Group of order 35 must have a normal subgroup of order 5 or 7
$begingroup$
Suppose $G$ is a group with identity element $1$, that $H$ and $K$ are normal subgroups of $G$, and that $Hcap K={1}$. Prove that if $h_{1},h_{2} in H$, $k_{1},k_{2} in K$ and $h_{1}k_{1}=h_{2}k_{2}$, then $h_{1}=h_{2}$ and $k_{1}=k_{2}$.
My attempt:
$h_{1}k_{1}=h_{2}k_{2}$ then $k_{1}=h_{1}^{-1}h_{2}k_{2}$
so $h_{1}^{-1}h_{2}=1$ and $k_{1}=k_{2}$, because $Hcap K={1}$.
Any tips, hints would be greatly appreciated.
group-theory
edited Mar 26 at 2:03
Shaun
11k113687
asked Mar 25 at 23:55
RivaldoRivaldo
236210
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a group with identity element $1$, that $H$ and $K$ are normal subgroups of $G$, and that $Hcap K={1}$. Prove that if $h_{1},h_{2} in H$, $k_{1},k_{2} in K$ and $h_{1}k_{1}=h_{2}k_{2}$, then $h_{1}=h_{2}$ and $k_{1}=k_{2}$.
My attempt:
$h_{1}k_{1}=h_{2}k_{2}$ then $k_{1}=h_{1}^{-1}h_{2}k_{2}$
so $h_{1}^{-1}h_{2}=1$ and $k_{1}=k_{2}$, because $Hcap K={1}$.
Any tips, hints would be greatly appreciated.
group-theory
edited Mar 26 at 2:03
Shaun
11k113687
asked Mar 25 at 23:55
RivaldoRivaldo
236210
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a group with identity element $1$, that $H$ and $K$ are normal subgroups of $G$, and that $Hcap K={1}$. Prove that if $h_{1},h_{2} in H$, $k_{1},k_{2} in K$ and $h_{1}k_{1}=h_{2}k_{2}$, then $h_{1}=h_{2}$ and $k_{1}=k_{2}$.
My attempt:
$h_{1}k_{1}=h_{2}k_{2}$ then $k_{1}=h_{1}^{-1}h_{2}k_{2}$
so $h_{1}^{-1}h_{2}=1$ and $k_{1}=k_{2}$, because $Hcap K={1}$.
Any tips, hints would be greatly appreciated.
group-theory
edited Mar 26 at 2:03
Shaun
11k113687
asked Mar 25 at 23:55
RivaldoRivaldo
236210
$endgroup$
Suppose $G$ is a group with identity element $1$, that $H$ and $K$ are normal subgroups of $G$, and that $Hcap K={1}$. Prove that if $h_{1},h_{2} in H$, $k_{1},k_{2} in K$ and $h_{1}k_{1}=h_{2}k_{2}$, then $h_{1}=h_{2}$ and $k_{1}=k_{2}$.
My attempt:
$h_{1}k_{1}=h_{2}k_{2}$ then $k_{1}=h_{1}^{-1}h_{2}k_{2}$
so $h_{1}^{-1}h_{2}=1$ and $k_{1}=k_{2}$, because $Hcap K={1}$.
Any tips, hints would be greatly appreciated.
group-theory
group-theory
edited Mar 26 at 2:03
Shaun
11k113687
asked Mar 25 at 23:55
RivaldoRivaldo
236210
edited Mar 26 at 2:03
Shaun
11k113687
asked Mar 25 at 23:55
RivaldoRivaldo
236210
edited Mar 26 at 2:03
Shaun
11k113687
edited Mar 26 at 2:03
Shaun
11k113687
edited Mar 26 at 2:03
Shaun
11k113687
11k113687
asked Mar 25 at 23:55
RivaldoRivaldo
236210
asked Mar 25 at 23:55
RivaldoRivaldo
236210
asked Mar 25 at 23:55
RivaldoRivaldo
236210
236210
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think you did not use the hypothesis correctly. $h_2^{-1}h_1=k_2k_1^{-1}$ so each side is in $H cap K={1}$ which gives $h_2=h_1$ and $k_2=k_1$
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
I think you did not use the hypothesis correctly. $h_2^{-1}h_1=k_2k_1^{-1}$ so each side is in $H cap K={1}$ which gives $h_2=h_1$ and $k_2=k_1$
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
add a comment |
$begingroup$
I think you did not use the hypothesis correctly. $h_2^{-1}h_1=k_2k_1^{-1}$ so each side is in $H cap K={1}$ which gives $h_2=h_1$ and $k_2=k_1$
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
add a comment |
$begingroup$
I think you did not use the hypothesis correctly. $h_2^{-1}h_1=k_2k_1^{-1}$ so each side is in $H cap K={1}$ which gives $h_2=h_1$ and $k_2=k_1$
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
$endgroup$
I think you did not use the hypothesis correctly. $h_2^{-1}h_1=k_2k_1^{-1}$ so each side is in $H cap K={1}$ which gives $h_2=h_1$ and $k_2=k_1$
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
answered Mar 25 at 23:59
Kavi Rama MurthyKavi Rama Murthy
76.4k53370
76.4k53370
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
add a comment |
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
Im trying to say that since $h_{1}^{-1}h_{2}$ is in H so it can't be in K unless they are both equal to the identity so the product must equal to 1 is that wrong?
$endgroup$
– Rivaldo
Mar 26 at 0:03
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
$begingroup$
@Rivaldo I know you have the right idea but you did not express it correctly.
$endgroup$
– Kavi Rama Murthy
Mar 26 at 0:05
add a comment |
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