Mdthbs

Why have both: BJT and FET transistors on IC output?

How do I safety check that there is no light in Darkroom / Darkbag?

Protect a 6 inch air hose from physical damage

Overprovisioning SSD on ubuntu. How? Ubuntu 19.04 Samsung SSD 860

Is it moral to remove/hide certain parts of a photo, as a photographer?

"Will flex for food". What does this phrase mean?

How to get maximum number that newcount can hold?

speaker impedence

UX writing: When to use "we"?

Gold Battle KoTH

What is the difference between 2/4 and 4/4 when it comes the accented beats?

Skipping same old introductions

Ernie and the Superconducting Boxes

Backpacking with incontinence

What sort of solar system / atmospheric conditions, if any, would allow for a very cold planet that still receives plenty of light from its sun?

How to draw twisted cuves?

Should I take up a Creative Writing online course?

Pre-Greek θάλασσα "thalassa" and Turkish talaz

Return last number in sub-sequences in a list of integers

What is the reason behind water not falling from a bucket at the top of loop?

What is the most 'environmentally friendly' way to learn to fly?

Why do we need a voltage divider when we get the same voltage at the output as the input?

How long should I wait to plug in my refrigerator after unplugging it?

If I buy and download a game through second Nintendo account do I own it on my main account too?

Look for an example where H is normal in K, K is characteristic in G, but H is not normal in G

An example for a calculation where imaginary numbers are used but don't occur in the question or the solution.$L : K$ is finite but not normal extensionExample of rings of the same positive characteristic that do not embed into their tensor product?an example for $pi$-quasinormal subgroup but is not normalLooking for a counter example - Normal subgroups and quotient groupGive an example of $ K subset H subset G$, such that $K triangleleft H$ and $H triangleleft G$ but $ K triangleleft G$ is not true.Example of a non-separable normal extensionExample for $|gN|<|g|$Duality Theorem. An example where the dual is feasible but the primal is notLooking for group $G$ such that $G$ nilpotent, $N leq G$, $N cap Z(G) = {1}$

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}

3

$begingroup$

Look for an example where $H triangleleft K triangleleft_{text{char}} G$ but $H ntriangleleft G$

Anyone got one? Thanks!!

abstract-algebra group-theory examples-counterexamples

share|cite|improve this question

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What does $H triangleleft K$ char $G$ mean?
  $endgroup$
  – lulu
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I assume it means "$H$ is normal in $K$ and $K$ is charateristic in $G$".
  $endgroup$
  – freakish
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Look for an example where $H triangleleft K triangleleft_{text{char}} G$ but $H ntriangleleft G$

Anyone got one? Thanks!!

abstract-algebra group-theory examples-counterexamples

share|cite|improve this question

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What does $H triangleleft K$ char $G$ mean?
  $endgroup$
  – lulu
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I assume it means "$H$ is normal in $K$ and $K$ is charateristic in $G$".
  $endgroup$
  – freakish
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Look for an example where $H triangleleft K triangleleft_{text{char}} G$ but $H ntriangleleft G$

Anyone got one? Thanks!!

abstract-algebra group-theory examples-counterexamples

share|cite|improve this question

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

$endgroup$

share|cite|improve this question

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

share|cite|improve this question

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

edited 2 hours ago

ThorWittich

3,81633 silver badges2121 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

asked 8 hours ago

Mathematical MushroomMathematical Mushroom

1,06222 silver badges1212 bronze badges

2 Answers
2

active

oldest

votes

4

$begingroup$

Take $G=A_5wr C_2$, which is a semidirect product of $A_5times A_5$ by $C_2$, with the action being an exchange of coordinates.

The commutator subgroup is $A_5times A_5$: it is clearly contained there, since the quotient is abelian. So the commutator subgroup must be a normal subgroup of $A_5times A_5$, and must contain the commutator subgroup of $A_5times A_5$. This is the whole group, so the commutator subgroup of $G$ is $A_5times A_5$, which is therefore characteristic (in fact, fully invariant) in $G$.

Now, $A_5times{e}$ is normal in $A_5times A_5$. However, it is not normal in $G$, since conjugation by the nontrivial element of $C_2$ maps it to ${e}times A_5$.

So take $G=A_5wr C_2$, $K=A_5times A_5$, $H=A_5times{e}$.

share|cite|improve this answer

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cool. Got any simpler examples?
  $endgroup$
  – Mathematical Mushroom
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @MathematicalMushroom: Given that this one involves the smallest simple group... in short: “simpler” is in the eye of the beholder, and without any parameters for measuring it, your request is nonsensical. This is a pretty basic construction, involving two very basic groups.
  $endgroup$
  – Arturo Magidin
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

5

$begingroup$

I can give you an example that at least is easier in my eyes (does not mean that you consider it to be easier as well). Consider the alternating group on $4$ elements $G = A_4$. Then we have $$H = langle (1 ; 2)(3 ; 4) rangle subset V_4 subset A_4, $$ where the first inclusion is an inclusion of a subgroup of index $2$, i.e. $H subset V_4$ is normal.

Why is $V_4 subset A_4$ characteristic?

The Klein four-group $V_4$ is given by all elements of order $2$ in $A_4$ (and the identity element) and automorphisms are order-preserving. Thus every automorphism of $A_4$ has to map $V_4$ to itself.

Now we have $$(1 ; 2 ; 3)^{-1}(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3 ; 2)(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3)(2 ; 4) notin H,$$ such that $H subset A_4$ is not a normal subgroup. Thus we have found an example.

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps a simpler way to note that $V$ is characteristic is to note that it is a normal Sylow subgroup.
  $endgroup$
  – Arturo Magidin
  9 mins ago
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3312561%2flook-for-an-example-where-h-is-normal-in-k-k-is-characteristic-in-g-but-h-is-n%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

Take $G=A_5wr C_2$, which is a semidirect product of $A_5times A_5$ by $C_2$, with the action being an exchange of coordinates.

The commutator subgroup is $A_5times A_5$: it is clearly contained there, since the quotient is abelian. So the commutator subgroup must be a normal subgroup of $A_5times A_5$, and must contain the commutator subgroup of $A_5times A_5$. This is the whole group, so the commutator subgroup of $G$ is $A_5times A_5$, which is therefore characteristic (in fact, fully invariant) in $G$.

Now, $A_5times{e}$ is normal in $A_5times A_5$. However, it is not normal in $G$, since conjugation by the nontrivial element of $C_2$ maps it to ${e}times A_5$.

So take $G=A_5wr C_2$, $K=A_5times A_5$, $H=A_5times{e}$.

share|cite|improve this answer

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cool. Got any simpler examples?
  $endgroup$
  – Mathematical Mushroom
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @MathematicalMushroom: Given that this one involves the smallest simple group... in short: “simpler” is in the eye of the beholder, and without any parameters for measuring it, your request is nonsensical. This is a pretty basic construction, involving two very basic groups.
  $endgroup$
  – Arturo Magidin
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Take $G=A_5wr C_2$, which is a semidirect product of $A_5times A_5$ by $C_2$, with the action being an exchange of coordinates.

The commutator subgroup is $A_5times A_5$: it is clearly contained there, since the quotient is abelian. So the commutator subgroup must be a normal subgroup of $A_5times A_5$, and must contain the commutator subgroup of $A_5times A_5$. This is the whole group, so the commutator subgroup of $G$ is $A_5times A_5$, which is therefore characteristic (in fact, fully invariant) in $G$.

Now, $A_5times{e}$ is normal in $A_5times A_5$. However, it is not normal in $G$, since conjugation by the nontrivial element of $C_2$ maps it to ${e}times A_5$.

So take $G=A_5wr C_2$, $K=A_5times A_5$, $H=A_5times{e}$.

share|cite|improve this answer

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Cool. Got any simpler examples?
  $endgroup$
  – Mathematical Mushroom
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  @MathematicalMushroom: Given that this one involves the smallest simple group... in short: “simpler” is in the eye of the beholder, and without any parameters for measuring it, your request is nonsensical. This is a pretty basic construction, involving two very basic groups.
  $endgroup$
  – Arturo Magidin
  5 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Take $G=A_5wr C_2$, which is a semidirect product of $A_5times A_5$ by $C_2$, with the action being an exchange of coordinates.

The commutator subgroup is $A_5times A_5$: it is clearly contained there, since the quotient is abelian. So the commutator subgroup must be a normal subgroup of $A_5times A_5$, and must contain the commutator subgroup of $A_5times A_5$. This is the whole group, so the commutator subgroup of $G$ is $A_5times A_5$, which is therefore characteristic (in fact, fully invariant) in $G$.

Now, $A_5times{e}$ is normal in $A_5times A_5$. However, it is not normal in $G$, since conjugation by the nontrivial element of $C_2$ maps it to ${e}times A_5$.

So take $G=A_5wr C_2$, $K=A_5times A_5$, $H=A_5times{e}$.

share|cite|improve this answer

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

answered 7 hours ago

Arturo MagidinArturo Magidin

273k3434 gold badges602602 silver badges935935 bronze badges

5

$begingroup$

I can give you an example that at least is easier in my eyes (does not mean that you consider it to be easier as well). Consider the alternating group on $4$ elements $G = A_4$. Then we have $$H = langle (1 ; 2)(3 ; 4) rangle subset V_4 subset A_4, $$ where the first inclusion is an inclusion of a subgroup of index $2$, i.e. $H subset V_4$ is normal.

Why is $V_4 subset A_4$ characteristic?

The Klein four-group $V_4$ is given by all elements of order $2$ in $A_4$ (and the identity element) and automorphisms are order-preserving. Thus every automorphism of $A_4$ has to map $V_4$ to itself.

Now we have $$(1 ; 2 ; 3)^{-1}(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3 ; 2)(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3)(2 ; 4) notin H,$$ such that $H subset A_4$ is not a normal subgroup. Thus we have found an example.

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps a simpler way to note that $V$ is characteristic is to note that it is a normal Sylow subgroup.
  $endgroup$
  – Arturo Magidin
  9 mins ago
  

  
  
  
  

add a comment | 

5

$begingroup$

I can give you an example that at least is easier in my eyes (does not mean that you consider it to be easier as well). Consider the alternating group on $4$ elements $G = A_4$. Then we have $$H = langle (1 ; 2)(3 ; 4) rangle subset V_4 subset A_4, $$ where the first inclusion is an inclusion of a subgroup of index $2$, i.e. $H subset V_4$ is normal.

Why is $V_4 subset A_4$ characteristic?

The Klein four-group $V_4$ is given by all elements of order $2$ in $A_4$ (and the identity element) and automorphisms are order-preserving. Thus every automorphism of $A_4$ has to map $V_4$ to itself.

Now we have $$(1 ; 2 ; 3)^{-1}(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3 ; 2)(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3)(2 ; 4) notin H,$$ such that $H subset A_4$ is not a normal subgroup. Thus we have found an example.

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Perhaps a simpler way to note that $V$ is characteristic is to note that it is a normal Sylow subgroup.
  $endgroup$
  – Arturo Magidin
  9 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

I can give you an example that at least is easier in my eyes (does not mean that you consider it to be easier as well). Consider the alternating group on $4$ elements $G = A_4$. Then we have $$H = langle (1 ; 2)(3 ; 4) rangle subset V_4 subset A_4, $$ where the first inclusion is an inclusion of a subgroup of index $2$, i.e. $H subset V_4$ is normal.

Why is $V_4 subset A_4$ characteristic?

The Klein four-group $V_4$ is given by all elements of order $2$ in $A_4$ (and the identity element) and automorphisms are order-preserving. Thus every automorphism of $A_4$ has to map $V_4$ to itself.

Now we have $$(1 ; 2 ; 3)^{-1}(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3 ; 2)(1 ; 2)(3 ; 4)(1 ; 2 ; 3) = (1 ; 3)(2 ; 4) notin H,$$ such that $H subset A_4$ is not a normal subgroup. Thus we have found an example.

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges

answered 5 hours ago

ThorWittichThorWittich

3,81633 silver badges2121 bronze badges