Linked Questions

15
votes
1answer
1k views

Does $gHg^{-1}\subseteq H$ imply $gHg^{-1}= H$? [duplicate]

Let $G$ be a group, $H<G$ a subgroup and $g$ an element of $G$. Let $\lambda_g$ denote the inner automorphism which maps $x$ to $gxg^{-1}$. I wonder if $H$ can be mapped to a proper subgroup of ...
8
votes
2answers
689 views

If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$? [duplicate]

If $G$ is a group, $H$ is a subgroup of $G$ and $g\in G$, is it possible that $gHg^{-1} \subset H$ ? This means, $gHg^{-1}$ is a proper subgroup of $H$. We know that $H \cong gHg^{-1}$, so if $H$ is ...
9
votes
2answers
624 views

If $gHg^{-1} \subset H$, must we have $g^{-1}Hg \subset H$? [duplicate]

Let $H$ be a subgroup of a group $G$. Consider the set $\{g \in G: gHg^{-1} \subset H\}$. Must this set always be a group? If $H$ was a finite subgroup then $gHg^{-1} \subset H$ if and only if $gHg^{-...
9
votes
1answer
727 views

Definition of the normalizer of a subgroup [duplicate]

Let $G$ be a group and $H$ a subgroup of $G$. Is there any counterexample to the assertion $N_G(H):=\{g\in G\mid gHg^{-1}=H\}=\{g\in G\mid gHg^{-1}\subset H\}$? Thanks!
3
votes
2answers
383 views

Example of a subgroup for normality [duplicate]

Give an example of a group $G$ and a subgroup $H$ of $G$, such that for some $g\in G,\ g^{-1}Hg\subset H$ i.e. $g^{-1}Hg$ is properly contained in $H$.
2
votes
2answers
105 views

Example of a group with non equal coset for a fixed element [duplicate]

Could anyone give me example of the following? A group $G$, a subgroup $H$, $a$ is a fixed element in $G$, such that $aH$ is a proper subset of $Ha$ and $aH \neq Ha$.
5
votes
1answer
179 views

An example of a group, a subgroup and an element, satisfying a given condition. [duplicate]

Is there a group $G$, a subgroup $H$ and an element $x$, such that $xHx^{-1} \subset H$, but $xHx^{-1} \neq H$? Thanks in advance.
4
votes
1answer
120 views

Alternative(?) definition for normalizer [duplicate]

Let $G$ be a group and $H$ be a subgroup. Consider the following set: $$\hat{N}(H):=\left\{g\in G: g^{-1}Hg \subset H\right\} $$ The normalizer of $H$ is usually defined as $$N(H):=\left\{g\in G: g^{-...
1
vote
1answer
102 views

Is Conjugate of Subgroup Could be Proper of its Subgroup? [duplicate]

I am curious whether it's possible for the group $G$ such that there is a subgroup $H ≤ G$ and an element $g ∈ G$ such that $gHg^{−1}$ $\subsetneq$ $H$ to exist. It means there exist some duplication ...
0
votes
1answer
99 views

Example of subgroup $H$ with the property $g H g^{-1}$ is properly contained in $H$. [duplicate]

After learning the definition of normal subgroup, I would like to find an example of a group $G$ which has a subgroup $H$ such that there exists an element $g \in G$ such that $gHg^{-1} \subsetneq H$, ...
0
votes
1answer
52 views

Normal subgroups of infinite groups [duplicate]

Suppose that $G$ is an infinite group and $H\le G$. The criterion for normal subgroups states that if $H^g\le H$ for all $g\in G$, then $H\trianglelefteq G$ (and thus $H^g=H$ for all $g\in G$). I am ...
0
votes
1answer
65 views

Finding a subgroup $H$ of $G$ where, for some $a \in G$, $aha^{-1} \in H$ but $a^{-1}ha \notin H$ (for all $h \in H$) [duplicate]

How can I prove the that the statements, (i) $aha^{-1} \in H$ for each $h \in H$ (ii) $a^{-1}ha \in H$ for each $h \in H$ are not equivalent for any particular $a \in G$? I have been trying to come ...
0
votes
0answers
78 views

Disprove for any subgroup $H$, we have $N(H)=\{g \in G\mid H\supseteq gHg^{-1}\}$. [duplicate]

Disprove for any subgroup $H$, we have $N(H)=\{g \in G\mid H\supseteq gHg^{-1}\}$, where $N(H)$ is the normalizer of $H$ in a group $G$ i.e. $N(H)=\{g \in G \mid H=gHg^{−1} \}$.
1
vote
0answers
58 views

Let $H$ be a subgroup of $G$. $S=\{g\in G\mid ghg^{-1}\in H\}$. Is $S= N_{G(H)}$? [duplicate]

Let $H$ be a subgroup of $G$. $S=\{g\in G\mid ghg^{-1}\in H,\forall h\in H\}$. I know that $N_G(H)\subset S$. But I can't seem to find any counterexamples for the other inclusion. Any ideas?
1
vote
1answer
67 views

Equivalent definition of normalizer of a group. Need help in proof. [duplicate]

The definition of normalizer of a non-empty set $A$ in group $G$ is given by $$N_G(A) = \{g\in G\:|\:gAg^{-1} = A\}$$ This definition is from Abstract Algebra by Dummit and Foote. Another definition ...

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