Is $\theta(N)$ a normal subgroup of $G$?

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In summary, a normal subgroup of a group G is a subgroup N that satisfies the property that for every element g in G, the conjugate of N by g, denoted gNg⁻¹, is also a subset of N. It is typically denoted by the symbol N with a vertical line through it, or by adding an "n" subscript to the subgroup symbol, such as N or Hn. A normal subgroup is important because it allows us to define a quotient group G/N, which is the set of all cosets of N in G. This allows for easier study of the group's structure. To determine if a subgroup is normal, we can use the definition or the criterion that if the index of the subgroup
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Ackbach
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Here is this week's POTW:

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Let $G$ be a group. If $\theta$ is an automorphism of $G$ and $N \vartriangleleft G$, prove that $\theta(N) \vartriangleleft G$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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This was Problem 2.5.27 on page 75 of I. N. Herstein's Abstract Algebra, 3rd Ed..

No one answered this week's POTW. The solution follows:

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By definition, $\theta$ is 1-1, onto, a homomorphism, and $\theta: G \to G$. By definition, since $N \vartriangleleft G$, it must be that $N$ is a subgroup of $G$, and $a^{-1}Na\in G$ for every $a\in G$. We first show that $\theta(N)$ is a subgroup of $G$. Since $\theta:G\to G$, and $N\subseteq G$, it follows that $\theta(N)\subseteq G$. Since $N$ is a subgroup, it follows that $e\in N$, and hence $\theta(e)=e\in G$, since homomorphisms map identities to identities (Lemma 2.5.2 in Herstein). Therefore, $\theta(N)\not=\varnothing$. Suppose $s, t\in\theta(N)$. By definition, there exist $m, n\in N$ such that $s=\theta(m)$ and $t=\theta(n)$. Then $st=\theta(m)\theta(n)=\theta(mn)$. Since $N$ is a subgroup, $mn\in N$, hence $st\in\theta(N)$, and $\theta(N)$ is closed under multiplication. Moreover, since $N$ is a subgroup, $m^{-1}\in N$, and hence $\theta(m^{-1})=\theta(m)^{-1}=s^{-1}\in\theta(N)$, since homomorphisms preserve inverses (Lemma 2.5.2 in Herstein, again). Since $\theta(N)$ is a nonempty subset of $G$, and is closed under multiplication and inverses, it is a subgroup.

Now let $a\in G$ be arbitrary, as well as $u\in\theta(N)$ be arbitrary. Then we wish to show that $aua^{-1}\in \theta(N)$. There exists $p\in N$ such that $u=\theta(p)$. Therefore, we have that $aua^{-1}=a \, \theta(p) \, a^{-1}=\theta(apa^{-1}).$ Since $N \vartriangleleft G$, it follows that $apa^{-1}\in N$, hence $\theta(apa^{-1}) \in \theta(N)$, and $\theta(N)\vartriangleleft G$, as required.
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FAQ: Is $\theta(N)$ a normal subgroup of $G$?

What is the definition of a normal subgroup?

A normal subgroup of a group G is a subgroup N that satisfies the property that for every element g in G, the conjugate of N by g, denoted gNg⁻¹, is also a subset of N.

How is a normal subgroup denoted in mathematical notation?

A normal subgroup is typically denoted by the symbol N with a vertical line through it, or by adding an "n" subscript to the subgroup symbol, such as N or Hn.

What is the significance of a normal subgroup?

A normal subgroup is important because it allows us to define a quotient group G/N, which is the set of all cosets of N in G. This allows us to study the structure of G in a more manageable way.

How do you determine if a subgroup is normal?

To determine if a subgroup is normal, we can use the definition and check if for every element g in G, gNg⁻¹ is a subset of N. Alternatively, we can use the criterion that if the index of the subgroup is equal to the index of its normalizer, then the subgroup is normal.

How does the normality of a subgroup affect the group's structure?

If a subgroup is normal, then the quotient group G/N inherits the same algebraic structure as G. This means that many properties and operations of G can also be applied to G/N, making it easier to study the group's structure.

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