Mixing
Let |G| = 100, $Hleq G$ with |H| = 25. If $gin G$ with $|g| = 5^k$ for some non-negative integer k, prove that $g in H$
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I'm not exactly too sure how to prove this. I know by Lagrange's theorem that G can have subgroups with orders, 1, 2, 5, 10, 20, 25, 50, 100. We're to assume that the group operation is multiplication.
Any help is appreciated
abstract-algebra group-theory
asked Jul 23 at 23:08
Wallace
625
add a comment |Â
up vote
3
down vote
favorite
I'm not exactly too sure how to prove this. I know by Lagrange's theorem that G can have subgroups with orders, 1, 2, 5, 10, 20, 25, 50, 100. We're to assume that the group operation is multiplication.
Any help is appreciated
abstract-algebra group-theory
asked Jul 23 at 23:08
Wallace
625
Let $g$ act on the cosets of $H$ by right multiplication. What does it mean for $g$ to fix a coset?
– Steve D
Jul 24 at 1:29
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I'm not exactly too sure how to prove this. I know by Lagrange's theorem that G can have subgroups with orders, 1, 2, 5, 10, 20, 25, 50, 100. We're to assume that the group operation is multiplication.
Any help is appreciated
abstract-algebra group-theory
asked Jul 23 at 23:08
Wallace
625
I'm not exactly too sure how to prove this. I know by Lagrange's theorem that G can have subgroups with orders, 1, 2, 5, 10, 20, 25, 50, 100. We're to assume that the group operation is multiplication.
Any help is appreciated
abstract-algebra group-theory
asked Jul 23 at 23:08
Wallace
625
asked Jul 23 at 23:08
Wallace
625
asked Jul 23 at 23:08
Wallace
625
asked Jul 23 at 23:08
Wallace
625
625
Let $g$ act on the cosets of $H$ by right multiplication. What does it mean for $g$ to fix a coset?
– Steve D
Jul 24 at 1:29
add a comment |Â
Let $g$ act on the cosets of $H$ by right multiplication. What does it mean for $g$ to fix a coset?
– Steve D
Jul 24 at 1:29
Let $g$ act on the cosets of $H$ by right multiplication. What does it mean for $g$ to fix a coset?
– Steve D
Jul 24 at 1:29
Let $g$ act on the cosets of $H$ by right multiplication. What does it mean for $g$ to fix a coset?
– Steve D
Jul 24 at 1:29
add a comment |Â
2 Answers
2
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oldest
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up vote
4
down vote
accepted
Here is an elementary approach. Suppose $gin G$ with $|g|=5^k$. By Lagrange's theorem, $5^k$ must divide $|G|=100$, so $k=0$, $1$ or $2$.
Case 1. $k=0$.
In this case, $g=e$ (the identity element), and so $gin H$.
Case 2. $k=1$. Let $K=langle g rangle = g, g^2, g^3, g^4, e$ be the subgroup generated by $g$. Consider the subset $Hcdot K = h cdot k: hin H text and kin K$ of $G$ (which is not necessarily a subgroup of $G$). Then it is an elementary fact that $|Hcdot K| = frac$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, and since $|K|=5$ is prime, we get $|Hcap K|=1$. But then
$|Hcdot K| = frac25cdot 51 = 125 > 100 = |G|$, a contradiction.
Case 3. $k=2$. This is similar to Case 2. Indeed, let $K=langle g rangle$ which has size $25$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, so by Lagrange's theorem (applied to $Hcap Ksubset K$), we get $|Hcap K|leq 5$. Again, we get $|Hcdot K|= frac geq frac25cdot 255 = 125 > 100 = |G|$, a contradiction.
answered Jul 24 at 1:22
Prism
5,22121773
add a comment |Â
up vote
1
down vote
Hint By Sylow Theorems $g$ belongs to a 5-Sylow subgroup and
$$n_5 equiv 1 pmod5 \
n_5|4$$
answered Jul 23 at 23:10
N. S.
97.7k5105197
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Here is an elementary approach. Suppose $gin G$ with $|g|=5^k$. By Lagrange's theorem, $5^k$ must divide $|G|=100$, so $k=0$, $1$ or $2$.
Case 1. $k=0$.
In this case, $g=e$ (the identity element), and so $gin H$.
Case 2. $k=1$. Let $K=langle g rangle = g, g^2, g^3, g^4, e$ be the subgroup generated by $g$. Consider the subset $Hcdot K = h cdot k: hin H text and kin K$ of $G$ (which is not necessarily a subgroup of $G$). Then it is an elementary fact that $|Hcdot K| = frac$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, and since $|K|=5$ is prime, we get $|Hcap K|=1$. But then
$|Hcdot K| = frac25cdot 51 = 125 > 100 = |G|$, a contradiction.
Case 3. $k=2$. This is similar to Case 2. Indeed, let $K=langle g rangle$ which has size $25$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, so by Lagrange's theorem (applied to $Hcap Ksubset K$), we get $|Hcap K|leq 5$. Again, we get $|Hcdot K|= frac geq frac25cdot 255 = 125 > 100 = |G|$, a contradiction.
answered Jul 24 at 1:22
Prism
5,22121773
add a comment |Â
up vote
4
down vote
accepted
Here is an elementary approach. Suppose $gin G$ with $|g|=5^k$. By Lagrange's theorem, $5^k$ must divide $|G|=100$, so $k=0$, $1$ or $2$.
Case 1. $k=0$.
In this case, $g=e$ (the identity element), and so $gin H$.
Case 2. $k=1$. Let $K=langle g rangle = g, g^2, g^3, g^4, e$ be the subgroup generated by $g$. Consider the subset $Hcdot K = h cdot k: hin H text and kin K$ of $G$ (which is not necessarily a subgroup of $G$). Then it is an elementary fact that $|Hcdot K| = frac$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, and since $|K|=5$ is prime, we get $|Hcap K|=1$. But then
$|Hcdot K| = frac25cdot 51 = 125 > 100 = |G|$, a contradiction.
Case 3. $k=2$. This is similar to Case 2. Indeed, let $K=langle g rangle$ which has size $25$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, so by Lagrange's theorem (applied to $Hcap Ksubset K$), we get $|Hcap K|leq 5$. Again, we get $|Hcdot K|= frac geq frac25cdot 255 = 125 > 100 = |G|$, a contradiction.
answered Jul 24 at 1:22
Prism
5,22121773
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Here is an elementary approach. Suppose $gin G$ with $|g|=5^k$. By Lagrange's theorem, $5^k$ must divide $|G|=100$, so $k=0$, $1$ or $2$.
Case 1. $k=0$.
In this case, $g=e$ (the identity element), and so $gin H$.
Case 2. $k=1$. Let $K=langle g rangle = g, g^2, g^3, g^4, e$ be the subgroup generated by $g$. Consider the subset $Hcdot K = h cdot k: hin H text and kin K$ of $G$ (which is not necessarily a subgroup of $G$). Then it is an elementary fact that $|Hcdot K| = frac$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, and since $|K|=5$ is prime, we get $|Hcap K|=1$. But then
$|Hcdot K| = frac25cdot 51 = 125 > 100 = |G|$, a contradiction.
Case 3. $k=2$. This is similar to Case 2. Indeed, let $K=langle g rangle$ which has size $25$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, so by Lagrange's theorem (applied to $Hcap Ksubset K$), we get $|Hcap K|leq 5$. Again, we get $|Hcdot K|= frac geq frac25cdot 255 = 125 > 100 = |G|$, a contradiction.
answered Jul 24 at 1:22
Prism
5,22121773
Here is an elementary approach. Suppose $gin G$ with $|g|=5^k$. By Lagrange's theorem, $5^k$ must divide $|G|=100$, so $k=0$, $1$ or $2$.
Case 1. $k=0$.
In this case, $g=e$ (the identity element), and so $gin H$.
Case 2. $k=1$. Let $K=langle g rangle = g, g^2, g^3, g^4, e$ be the subgroup generated by $g$. Consider the subset $Hcdot K = h cdot k: hin H text and kin K$ of $G$ (which is not necessarily a subgroup of $G$). Then it is an elementary fact that $|Hcdot K| = frac$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, and since $|K|=5$ is prime, we get $|Hcap K|=1$. But then
$|Hcdot K| = frac25cdot 51 = 125 > 100 = |G|$, a contradiction.
Case 3. $k=2$. This is similar to Case 2. Indeed, let $K=langle g rangle$ which has size $25$. If $gnotin H$, then $Hcap K$ is a proper subgroup of $K$, so by Lagrange's theorem (applied to $Hcap Ksubset K$), we get $|Hcap K|leq 5$. Again, we get $|Hcdot K|= frac geq frac25cdot 255 = 125 > 100 = |G|$, a contradiction.
answered Jul 24 at 1:22
Prism
5,22121773
answered Jul 24 at 1:22
Prism
5,22121773
answered Jul 24 at 1:22
Prism
5,22121773
answered Jul 24 at 1:22
Prism
5,22121773
5,22121773
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint By Sylow Theorems $g$ belongs to a 5-Sylow subgroup and
$$n_5 equiv 1 pmod5 \
n_5|4$$
answered Jul 23 at 23:10
N. S.
97.7k5105197
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
add a comment |Â
up vote
1
down vote
Hint By Sylow Theorems $g$ belongs to a 5-Sylow subgroup and
$$n_5 equiv 1 pmod5 \
n_5|4$$
answered Jul 23 at 23:10
N. S.
97.7k5105197
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint By Sylow Theorems $g$ belongs to a 5-Sylow subgroup and
$$n_5 equiv 1 pmod5 \
n_5|4$$
answered Jul 23 at 23:10
N. S.
97.7k5105197
Hint By Sylow Theorems $g$ belongs to a 5-Sylow subgroup and
$$n_5 equiv 1 pmod5 \
n_5|4$$
answered Jul 23 at 23:10
N. S.
97.7k5105197
answered Jul 23 at 23:10
N. S.
97.7k5105197
answered Jul 23 at 23:10
N. S.
97.7k5105197
answered Jul 23 at 23:10
N. S.
97.7k5105197
97.7k5105197
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
add a comment |Â
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
This seems a bit too "advanced" for the topics we've covered this far in our course. Is there perhaps a way you could attempt this using Lagrange's theorem? We have yet to cover "Sylow Theorem" so your solution doesn't make much sense to me
– Wallace
Jul 23 at 23:30
add a comment |Â
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Let $g$ act on the cosets of $H$ by right multiplication. What does it mean for $g$ to fix a coset?
– Steve D
Jul 24 at 1:29