Is $U_8$ isomorphic to $K_4$ ( Klein Group)
$begingroup$
$U_8=1,3,5,7$ since this group has one element of order one, three elements of two order and no element of $4$ order .. so does the Klein group.
Both $U(8)$ and the Klein group are non cyclic groups whose every proper subgroup is cyclic, so the Klein group is isomorphic to U(8)?
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 26 '18 at 13:31
amWhy
1
asked Dec 26 '18 at 13:08
HenryHenry
327
$endgroup$
add a comment |
$begingroup$
$U_8=1,3,5,7$ since this group has one element of order one, three elements of two order and no element of $4$ order .. so does the Klein group.
Both $U(8)$ and the Klein group are non cyclic groups whose every proper subgroup is cyclic, so the Klein group is isomorphic to U(8)?
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 26 '18 at 13:31
amWhy
1
asked Dec 26 '18 at 13:08
HenryHenry
327
$endgroup$
$begingroup$
is this the isomorphism defined](i.stack.imgur.com/HZTuj.jpg)](https://i.stack.imgur.com/…
$endgroup$
– Henry
Dec 26 '18 at 13:10
$begingroup$
See also this question, and this one.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 15:51
add a comment |
$begingroup$
$U_8=1,3,5,7$ since this group has one element of order one, three elements of two order and no element of $4$ order .. so does the Klein group.
Both $U(8)$ and the Klein group are non cyclic groups whose every proper subgroup is cyclic, so the Klein group is isomorphic to U(8)?
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 26 '18 at 13:31
amWhy
1
asked Dec 26 '18 at 13:08
HenryHenry
327
$endgroup$
$U_8=1,3,5,7$ since this group has one element of order one, three elements of two order and no element of $4$ order .. so does the Klein group.
Both $U(8)$ and the Klein group are non cyclic groups whose every proper subgroup is cyclic, so the Klein group is isomorphic to U(8)?
abstract-algebra group-theory finite-groups group-isomorphism
abstract-algebra group-theory finite-groups group-isomorphism
edited Dec 26 '18 at 13:31
amWhy
1
asked Dec 26 '18 at 13:08
HenryHenry
327
edited Dec 26 '18 at 13:31
amWhy
1
asked Dec 26 '18 at 13:08
HenryHenry
327
edited Dec 26 '18 at 13:31
amWhy
1
edited Dec 26 '18 at 13:31
amWhy
1
edited Dec 26 '18 at 13:31
amWhy
1
1
asked Dec 26 '18 at 13:08
HenryHenry
327
asked Dec 26 '18 at 13:08
HenryHenry
327
asked Dec 26 '18 at 13:08
HenryHenry
327
327
$begingroup$
is this the isomorphism defined](i.stack.imgur.com/HZTuj.jpg)](https://i.stack.imgur.com/…
$endgroup$
– Henry
Dec 26 '18 at 13:10
$begingroup$
See also this question, and this one.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 15:51
add a comment |
$begingroup$
is this the isomorphism defined](i.stack.imgur.com/HZTuj.jpg)](https://i.stack.imgur.com/…
$endgroup$
– Henry
Dec 26 '18 at 13:10
$begingroup$
See also this question, and this one.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 15:51
$begingroup$
is this the isomorphism defined](i.stack.imgur.com/HZTuj.jpg)](https://i.stack.imgur.com/…
$endgroup$
– Henry
Dec 26 '18 at 13:10
$begingroup$
is this the isomorphism defined](i.stack.imgur.com/HZTuj.jpg)](https://i.stack.imgur.com/…
$endgroup$
– Henry
Dec 26 '18 at 13:10
$begingroup$
See also this question, and this one.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 15:51
$begingroup$
See also this question, and this one.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 15:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There is only two groups of order four: (1) the cyclic group and (2) the Klein group.
As all elements of $U(8)$ are of order $2$, $U(8)$ is indeed isomorphic as a group to the Klein group.
The key argument is that there is no other groups of order four than the two mentioned above
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
add a comment |
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$begingroup$
There is only two groups of order four: (1) the cyclic group and (2) the Klein group.
As all elements of $U(8)$ are of order $2$, $U(8)$ is indeed isomorphic as a group to the Klein group.
The key argument is that there is no other groups of order four than the two mentioned above
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
add a comment |
$begingroup$
There is only two groups of order four: (1) the cyclic group and (2) the Klein group.
As all elements of $U(8)$ are of order $2$, $U(8)$ is indeed isomorphic as a group to the Klein group.
The key argument is that there is no other groups of order four than the two mentioned above
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
add a comment |
$begingroup$
There is only two groups of order four: (1) the cyclic group and (2) the Klein group.
As all elements of $U(8)$ are of order $2$, $U(8)$ is indeed isomorphic as a group to the Klein group.
The key argument is that there is no other groups of order four than the two mentioned above
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
$endgroup$
There is only two groups of order four: (1) the cyclic group and (2) the Klein group.
As all elements of $U(8)$ are of order $2$, $U(8)$ is indeed isomorphic as a group to the Klein group.
The key argument is that there is no other groups of order four than the two mentioned above
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
answered Dec 26 '18 at 13:13
mathcounterexamples.netmathcounterexamples.net
26.8k22157
26.8k22157
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
add a comment |
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
hey can u help finding all proper subgroups of Z2 × Z2 × Z2 ? i want to know how to approach to such problems ..any methodol
$endgroup$
– Henry
Dec 26 '18 at 13:27
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
$begingroup$
I suggest you post another question to avoid mixing different topics.
$endgroup$
– mathcounterexamples.net
Dec 26 '18 at 13:28
add a comment |
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$begingroup$
is this the isomorphism defined](i.stack.imgur.com/HZTuj.jpg)](https://i.stack.imgur.com/…
$endgroup$
– Henry
Dec 26 '18 at 13:10
$begingroup$
See also this question, and this one.
$endgroup$
– Dietrich Burde
Dec 26 '18 at 15:51