Prove the existence of order 4 subgroups of order 8 groups
$begingroup$
I am participating in an Introductory course in groups and I have the following question:
Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.
i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
group-theory finite-groups
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
$endgroup$
add a comment |
$begingroup$
I am participating in an Introductory course in groups and I have the following question:
Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.
i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
group-theory finite-groups
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
$endgroup$
$begingroup$
Do you know Lagrange's theorem ?
$endgroup$
– Tlön Uqbar Orbis Tertius
May 2 '15 at 9:01
$begingroup$
If you are familiar with Sylow theorem, and $p$-groups there is an even more general statement: If $p^k$ divides the order of $G$, where $p$ is prime, then $G$ has a subgroup of order $p^k$. In our case we have $2^2$ divides $8$. Of course this exercise is trivial with Sylow, so I am guessing you are not familiar, but it is sometimes nice to be aware of more general results.
$endgroup$
– Paul Plummer
May 2 '15 at 9:11
$begingroup$
Lagrange's theorem is the first thing, you will learn about finite groups. There is almost no chance to solve an exercise in abstract finite group theory without Lagrange's theorem.
$endgroup$
– MooS
May 2 '15 at 9:11
add a comment |
$begingroup$
I am participating in an Introductory course in groups and I have the following question:
Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.
i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
group-theory finite-groups
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
$endgroup$
I am participating in an Introductory course in groups and I have the following question:
Let $G$ be a finite group of order $8$. Prove that $G$ has a subgroup of order $4$ and a subgroup of order $2$.
i know how to prove that $G$ has a subgroup of order $2$ but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.
group-theory finite-groups
group-theory finite-groups
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
edited May 2 '15 at 9:29
Paul Plummer
5,41622051
5,41622051
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
asked May 2 '15 at 8:58
dorsh605dorsh605
553213
553213
$begingroup$
Do you know Lagrange's theorem ?
$endgroup$
– Tlön Uqbar Orbis Tertius
May 2 '15 at 9:01
$begingroup$
If you are familiar with Sylow theorem, and $p$-groups there is an even more general statement: If $p^k$ divides the order of $G$, where $p$ is prime, then $G$ has a subgroup of order $p^k$. In our case we have $2^2$ divides $8$. Of course this exercise is trivial with Sylow, so I am guessing you are not familiar, but it is sometimes nice to be aware of more general results.
$endgroup$
– Paul Plummer
May 2 '15 at 9:11
$begingroup$
Lagrange's theorem is the first thing, you will learn about finite groups. There is almost no chance to solve an exercise in abstract finite group theory without Lagrange's theorem.
$endgroup$
– MooS
May 2 '15 at 9:11
add a comment |
$begingroup$
Do you know Lagrange's theorem ?
$endgroup$
– Tlön Uqbar Orbis Tertius
May 2 '15 at 9:01
$begingroup$
If you are familiar with Sylow theorem, and $p$-groups there is an even more general statement: If $p^k$ divides the order of $G$, where $p$ is prime, then $G$ has a subgroup of order $p^k$. In our case we have $2^2$ divides $8$. Of course this exercise is trivial with Sylow, so I am guessing you are not familiar, but it is sometimes nice to be aware of more general results.
$endgroup$
– Paul Plummer
May 2 '15 at 9:11
$begingroup$
Lagrange's theorem is the first thing, you will learn about finite groups. There is almost no chance to solve an exercise in abstract finite group theory without Lagrange's theorem.
$endgroup$
– MooS
May 2 '15 at 9:11
$begingroup$
Do you know Lagrange's theorem ?
$endgroup$
– Tlön Uqbar Orbis Tertius
May 2 '15 at 9:01
$begingroup$
Do you know Lagrange's theorem ?
$endgroup$
– Tlön Uqbar Orbis Tertius
May 2 '15 at 9:01
$begingroup$
If you are familiar with Sylow theorem, and $p$-groups there is an even more general statement: If $p^k$ divides the order of $G$, where $p$ is prime, then $G$ has a subgroup of order $p^k$. In our case we have $2^2$ divides $8$. Of course this exercise is trivial with Sylow, so I am guessing you are not familiar, but it is sometimes nice to be aware of more general results.
$endgroup$
– Paul Plummer
May 2 '15 at 9:11
$begingroup$
If you are familiar with Sylow theorem, and $p$-groups there is an even more general statement: If $p^k$ divides the order of $G$, where $p$ is prime, then $G$ has a subgroup of order $p^k$. In our case we have $2^2$ divides $8$. Of course this exercise is trivial with Sylow, so I am guessing you are not familiar, but it is sometimes nice to be aware of more general results.
$endgroup$
– Paul Plummer
May 2 '15 at 9:11
$begingroup$
Lagrange's theorem is the first thing, you will learn about finite groups. There is almost no chance to solve an exercise in abstract finite group theory without Lagrange's theorem.
$endgroup$
– MooS
May 2 '15 at 9:11
$begingroup$
Lagrange's theorem is the first thing, you will learn about finite groups. There is almost no chance to solve an exercise in abstract finite group theory without Lagrange's theorem.
$endgroup$
– MooS
May 2 '15 at 9:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $G$ is cyclic, that is, if $G={1,x,x^2,x^3,x^4,x^5,x^6,x^7}$, then the elements with even exponents form a subgroup of order $4$.
If not, then take any element $xneq 1$. If its order is $4$, then we are done (the subgroup would be ${1,x,x^2,x^3}$, and if its order is $2$, take any other element $y$.
If the order of $y$ is $4$, then we are done, just like before. If the order of $y$ is $2$, then consider the subset $H={1,x,y,xy}$. If the order of $xy$ is $2$, then $H$ is a subgroup of order $4$. If the order of $xy$ is $4$, it generates a subgroup of order $4$.
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
$endgroup$
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
add a comment |
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$begingroup$
If $G$ is cyclic, that is, if $G={1,x,x^2,x^3,x^4,x^5,x^6,x^7}$, then the elements with even exponents form a subgroup of order $4$.
If not, then take any element $xneq 1$. If its order is $4$, then we are done (the subgroup would be ${1,x,x^2,x^3}$, and if its order is $2$, take any other element $y$.
If the order of $y$ is $4$, then we are done, just like before. If the order of $y$ is $2$, then consider the subset $H={1,x,y,xy}$. If the order of $xy$ is $2$, then $H$ is a subgroup of order $4$. If the order of $xy$ is $4$, it generates a subgroup of order $4$.
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
$endgroup$
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
add a comment |
$begingroup$
If $G$ is cyclic, that is, if $G={1,x,x^2,x^3,x^4,x^5,x^6,x^7}$, then the elements with even exponents form a subgroup of order $4$.
If not, then take any element $xneq 1$. If its order is $4$, then we are done (the subgroup would be ${1,x,x^2,x^3}$, and if its order is $2$, take any other element $y$.
If the order of $y$ is $4$, then we are done, just like before. If the order of $y$ is $2$, then consider the subset $H={1,x,y,xy}$. If the order of $xy$ is $2$, then $H$ is a subgroup of order $4$. If the order of $xy$ is $4$, it generates a subgroup of order $4$.
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
$endgroup$
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
add a comment |
$begingroup$
If $G$ is cyclic, that is, if $G={1,x,x^2,x^3,x^4,x^5,x^6,x^7}$, then the elements with even exponents form a subgroup of order $4$.
If not, then take any element $xneq 1$. If its order is $4$, then we are done (the subgroup would be ${1,x,x^2,x^3}$, and if its order is $2$, take any other element $y$.
If the order of $y$ is $4$, then we are done, just like before. If the order of $y$ is $2$, then consider the subset $H={1,x,y,xy}$. If the order of $xy$ is $2$, then $H$ is a subgroup of order $4$. If the order of $xy$ is $4$, it generates a subgroup of order $4$.
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
$endgroup$
If $G$ is cyclic, that is, if $G={1,x,x^2,x^3,x^4,x^5,x^6,x^7}$, then the elements with even exponents form a subgroup of order $4$.
If not, then take any element $xneq 1$. If its order is $4$, then we are done (the subgroup would be ${1,x,x^2,x^3}$, and if its order is $2$, take any other element $y$.
If the order of $y$ is $4$, then we are done, just like before. If the order of $y$ is $2$, then consider the subset $H={1,x,y,xy}$. If the order of $xy$ is $2$, then $H$ is a subgroup of order $4$. If the order of $xy$ is $4$, it generates a subgroup of order $4$.
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
edited May 2 '15 at 9:21
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
answered May 2 '15 at 9:05
ajotatxeajotatxe
54.3k24090
54.3k24090
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
add a comment |
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
how do i know that there is an element $ynot =1$ of order 2 other than the element $x$ i have already found?
$endgroup$
– dorsh605
May 2 '15 at 9:12
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
The order of every element divides the order of the group. This is a consequence of Lagrange's theorem.
$endgroup$
– ajotatxe
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
I meant, why can't all the other elements have order 8?
$endgroup$
– dorsh605
May 2 '15 at 9:15
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
There is one thing missing, and that is: One needs to have that ${1, x,, y , xy}$ is a subgroup, which seems to me still non-obvious. But it is easily fixable.
$endgroup$
– Pavel Čoupek
May 2 '15 at 9:16
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
$begingroup$
@dorsh605 Well, the unit has always order $1$. And if $x$ has order $8$, then $x^2$ has order $4$.
$endgroup$
– ajotatxe
May 2 '15 at 9:17
add a comment |
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$begingroup$
Do you know Lagrange's theorem ?
$endgroup$
– Tlön Uqbar Orbis Tertius
May 2 '15 at 9:01
$begingroup$
If you are familiar with Sylow theorem, and $p$-groups there is an even more general statement: If $p^k$ divides the order of $G$, where $p$ is prime, then $G$ has a subgroup of order $p^k$. In our case we have $2^2$ divides $8$. Of course this exercise is trivial with Sylow, so I am guessing you are not familiar, but it is sometimes nice to be aware of more general results.
$endgroup$
– Paul Plummer
May 2 '15 at 9:11
$begingroup$
Lagrange's theorem is the first thing, you will learn about finite groups. There is almost no chance to solve an exercise in abstract finite group theory without Lagrange's theorem.
$endgroup$
– MooS
May 2 '15 at 9:11