Prove the existence of order 4 subgroups of order 8 groups












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 2 '15 at 9:29









Paul Plummer

5,41622051




5,41622051










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


















  • $begingroup$
    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










1 Answer
1






active

oldest

votes


















5












$begingroup$

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 $4$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














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1 Answer
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1 Answer
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active

oldest

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active

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active

oldest

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5












$begingroup$

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 $4$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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


















5












$begingroup$

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 $4$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















5












5








5





$begingroup$

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 $4$.






share|cite|improve this answer











$endgroup$



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 $4$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited May 2 '15 at 9:21

























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




















  • $begingroup$
    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




















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