Relationship between prime power and the divisor
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Background
When trying to prove that a group G has a non-trivial centre if its order is a power of a prime $p$, there involves a step in which we claim the number of left cosets of a centraliser subgroup is also a power of $p$.
Problem
Let $q|p^k$, then why are we sure that $q = p^i$ where $0leq i <k$? In other words, why are we sure that $q$ can only be a power of $p$?
group-theory number-theory sylow-theory
asked 7 hours ago
hephaes
1167
add a comment |
up vote
0
down vote
favorite
Background
When trying to prove that a group G has a non-trivial centre if its order is a power of a prime $p$, there involves a step in which we claim the number of left cosets of a centraliser subgroup is also a power of $p$.
Problem
Let $q|p^k$, then why are we sure that $q = p^i$ where $0leq i <k$? In other words, why are we sure that $q$ can only be a power of $p$?
group-theory number-theory sylow-theory
asked 7 hours ago
hephaes
1167
If it weren't, it wouldn't divide $p$ since $p$ is prime...
– Rushabh Mehta
7 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Background
When trying to prove that a group G has a non-trivial centre if its order is a power of a prime $p$, there involves a step in which we claim the number of left cosets of a centraliser subgroup is also a power of $p$.
Problem
Let $q|p^k$, then why are we sure that $q = p^i$ where $0leq i <k$? In other words, why are we sure that $q$ can only be a power of $p$?
group-theory number-theory sylow-theory
asked 7 hours ago
hephaes
1167
Background
When trying to prove that a group G has a non-trivial centre if its order is a power of a prime $p$, there involves a step in which we claim the number of left cosets of a centraliser subgroup is also a power of $p$.
Problem
Let $q|p^k$, then why are we sure that $q = p^i$ where $0leq i <k$? In other words, why are we sure that $q$ can only be a power of $p$?
group-theory number-theory sylow-theory
group-theory number-theory sylow-theory
asked 7 hours ago
hephaes
1167
asked 7 hours ago
hephaes
1167
asked 7 hours ago
hephaes
1167
asked 7 hours ago
hephaes
1167
asked 7 hours ago
hephaes
1167
1167
If it weren't, it wouldn't divide $p$ since $p$ is prime...
– Rushabh Mehta
7 hours ago
add a comment |
If it weren't, it wouldn't divide $p$ since $p$ is prime...
– Rushabh Mehta
7 hours ago
If it weren't, it wouldn't divide $p$ since $p$ is prime...
– Rushabh Mehta
7 hours ago
If it weren't, it wouldn't divide $p$ since $p$ is prime...
– Rushabh Mehta
7 hours ago
add a comment |
1 Answer
1
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up vote
2
down vote
accepted
$$ q|p^k iff p^k bmod q =0 iff p^k=d cdot q$$
However, $d$ can't be anything else but:
$$d=p^lambda, quad 0<lambdaleq k $$
because $p$ is a prime ($p$'s powers have only divisors $1$ and other $p$'s powers) and thus
$$p^k=p^lambda cdot q iff q=p^{k-lambda}=p^i, quad 0 leq i<k$$
answered 7 hours ago
Jevaut
4639
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
$$ q|p^k iff p^k bmod q =0 iff p^k=d cdot q$$
However, $d$ can't be anything else but:
$$d=p^lambda, quad 0<lambdaleq k $$
because $p$ is a prime ($p$'s powers have only divisors $1$ and other $p$'s powers) and thus
$$p^k=p^lambda cdot q iff q=p^{k-lambda}=p^i, quad 0 leq i<k$$
answered 7 hours ago
Jevaut
4639
add a comment |
up vote
2
down vote
accepted
$$ q|p^k iff p^k bmod q =0 iff p^k=d cdot q$$
However, $d$ can't be anything else but:
$$d=p^lambda, quad 0<lambdaleq k $$
because $p$ is a prime ($p$'s powers have only divisors $1$ and other $p$'s powers) and thus
$$p^k=p^lambda cdot q iff q=p^{k-lambda}=p^i, quad 0 leq i<k$$
answered 7 hours ago
Jevaut
4639
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
$$ q|p^k iff p^k bmod q =0 iff p^k=d cdot q$$
However, $d$ can't be anything else but:
$$d=p^lambda, quad 0<lambdaleq k $$
because $p$ is a prime ($p$'s powers have only divisors $1$ and other $p$'s powers) and thus
$$p^k=p^lambda cdot q iff q=p^{k-lambda}=p^i, quad 0 leq i<k$$
answered 7 hours ago
Jevaut
4639
$$ q|p^k iff p^k bmod q =0 iff p^k=d cdot q$$
However, $d$ can't be anything else but:
$$d=p^lambda, quad 0<lambdaleq k $$
because $p$ is a prime ($p$'s powers have only divisors $1$ and other $p$'s powers) and thus
$$p^k=p^lambda cdot q iff q=p^{k-lambda}=p^i, quad 0 leq i<k$$
answered 7 hours ago
Jevaut
4639
edited 7 hours ago
answered 7 hours ago
Jevaut
4639
answered 7 hours ago
Jevaut
4639
answered 7 hours ago
Jevaut
4639
4639
add a comment |
add a comment |
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If it weren't, it wouldn't divide $p$ since $p$ is prime...
– Rushabh Mehta
7 hours ago