Are all groups of order 175 abelian?
$begingroup$
Question. Are all groups of order 175 abelian?
I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.
I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.
Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.
Could somebody please explain:
Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?
Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?
Anything else I should know?
Thanks
abstract-algebra group-theory finite-groups
edited Nov 30 '18 at 23:37
the_fox
2,58711533
asked May 21 '12 at 11:03
rk101rk101
12829
$endgroup$
add a comment |
$begingroup$
Question. Are all groups of order 175 abelian?
I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.
I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.
Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.
Could somebody please explain:
Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?
Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?
Anything else I should know?
Thanks
abstract-algebra group-theory finite-groups
edited Nov 30 '18 at 23:37
the_fox
2,58711533
asked May 21 '12 at 11:03
rk101rk101
12829
$endgroup$
$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07
$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10
$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12
3
$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16
1
$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28
add a comment |
$begingroup$
Question. Are all groups of order 175 abelian?
I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.
I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.
Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.
Could somebody please explain:
Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?
Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?
Anything else I should know?
Thanks
abstract-algebra group-theory finite-groups
edited Nov 30 '18 at 23:37
the_fox
2,58711533
asked May 21 '12 at 11:03
rk101rk101
12829
$endgroup$
Question. Are all groups of order 175 abelian?
I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.
I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| cdot |K|$.
Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.
Could somebody please explain:
Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?
Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?
Anything else I should know?
Thanks
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited Nov 30 '18 at 23:37
the_fox
2,58711533
asked May 21 '12 at 11:03
rk101rk101
12829
edited Nov 30 '18 at 23:37
the_fox
2,58711533
asked May 21 '12 at 11:03
rk101rk101
12829
edited Nov 30 '18 at 23:37
the_fox
2,58711533
edited Nov 30 '18 at 23:37
the_fox
2,58711533
edited Nov 30 '18 at 23:37
the_fox
2,58711533
2,58711533
asked May 21 '12 at 11:03
rk101rk101
12829
asked May 21 '12 at 11:03
rk101rk101
12829
asked May 21 '12 at 11:03
rk101rk101
12829
12829
$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07
$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10
$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12
3
$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16
1
$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28
add a comment |
$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07
$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10
$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12
3
$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16
1
$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28
$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07
$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07
$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10
$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10
$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12
$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12
3
3
$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16
$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16
1
1
$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28
$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
$endgroup$
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
$endgroup$
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
add a comment |
$begingroup$
Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
$endgroup$
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
add a comment |
$begingroup$
Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
$endgroup$
Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
answered May 21 '12 at 11:12
DonAntonioDonAntonio
177k1492227
177k1492227
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
add a comment |
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Why do we need to show they're normal?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:21
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
$begingroup$
Is it because we want to show that $G$ is a direct product?
$endgroup$
– TheLast Cipher
Sep 26 '18 at 11:54
add a comment |
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$begingroup$
Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G cong H times K$ is abelian.
$endgroup$
– martini
May 21 '12 at 11:07
$begingroup$
Only because they intersect trivially?
$endgroup$
– rk101
May 21 '12 at 11:10
$begingroup$
For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, bbc.co.uk/news and tearfund.org)
$endgroup$
– user1729
May 21 '12 at 11:12
3
$begingroup$
(Also, for (1), let $G$ be a group where $G=HK$ and $Hcap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=Htimes K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $Hrtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=Hbowtie K$ (see my answer here math.stackexchange.com/questions/107781/…)
$endgroup$
– user1729
May 21 '12 at 11:16
1
$begingroup$
@rk101, see also math.stackexchange.com/questions/67407/…
$endgroup$
– Nicky Hekster
May 21 '12 at 22:28