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My question reads:

Let K be a subgroup of a group G index 2. Prove that N is a normal subgroup as follows:

a) If a is not in N, prove that the coset Na consists of all the elements of G that are not in N

b) For each a in G, prove that a^(-1)Na ⊆ N

I do not understand how to approach this problem, for part a since a is not in N then is it not just straightforward that Na does not have elements in N? I am not too sure where to proof part here begins.

I would like help on getting this proof started so I can finish it.

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    You are correct, $N$ and $Na$ are disjoint. Why must their union be all of $G$? – 2017-02-21
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    @BobJones they are disjoint because a is not in N? – 2017-02-21
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    Yeah. What would happen if you had $h_1a=h_2$ for $h_1, h_2\in N$? – 2017-02-21
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    @BobJones then a is in N as well? – 2017-02-21
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    Yes, contradiction. So I ask again: why must their union be all of G? – 2017-02-21
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    @BobJones because we are just considering what is in N and what is not – 2017-02-21
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    But why is everything not in $N$ equal to something in $N$ times $a$? – 2017-02-21
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    @BobJones because is its not in N it has to be in Na because it is the only other option – 2017-02-21
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    You're using the "index 2" part of the question, correct? I didn't get that from your answers, but if so, then yes, the question is finished. – 2017-02-21
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    @BobJones yes since it is index 2 there are only 2 distinct right cosets of N in G – 2017-02-21

1 Answers 1

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We'll do it procedurally. Do ask questions if you are dissatisfied.

So, first in part a, given $a \notin N$, we want to show that $Na$ consists of all elements that are not in $N$.

To do this, we have to show two things:

1) Every element in $Na$ is not in $N$. To do this, we let $x \in Na$, then $x = na$ for some $n \in N$.Suppose, by way of contradiction, we assume $x \in N$. Then, right-multiplying both sides by $n^{-1}$, we get $a = xn^{-1}$. Now, $x \in N$, $n^{-1} \in N$ as $N$ is a subgroup, so $a \in N$, giving a contradiction. Hence, $x \notin N$, so $Na$ contains elements which at least,are not in $N$.

2) Every element not in $N$, is in $Na$. To do this, note that $N$ is of index $2$ in $G$. Hence, the number of cosets of $N$ is $2$. Can we describe these cosets? One coset is $N$ itself (or $Ne$, if you like). The other coset, is just described by taking $a \notin N$, and forming $Na$. Because $N$ is of index two, these are the only two cosets. Since every element of $G$ must belong to a coset (by definition of cosets being a partition by equivalence classes of $G$), we have that $Na$ contains exactly all elements that are not in $N$.

Now, we can go to part $b$:Consider $aNa^{-1}$. If $a \in N$, then this just reduces to $N$, so $a^{-1}Na \subset N$.

If $a \notin N$, then note that $Na$ consists of all elements not in $N$. Similarly, $aN$ consists of all elements not in $N$ (same argument for left cosets). By the description of these sets, $aN = Na$, or $aNa^{-1} = N$. Taking inverse of both sides, $a^{-1}Na = N$.

Hence, this completes the proof that $N$ is a normal subgroup of $G$.

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    For your part 2) shouldn't it read Na? – 2017-02-21
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    @Sam Thank you for the correction.However, are you satisfied with the proof? We can clarify things right here. – 2017-02-21
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    for part b, can we take an element a^(-1)na and say it is either in N or Na by part a? – 2017-02-21
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    Yes, we can do that... oh, that's a wonderful observation, that really finishes the proof in fact! – 2017-02-21
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    Okay, so how would this change the proof instead for b if we consider a not in N and n in N where a^(-1)na is either in N or Na? – 2017-02-21
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    @Sam Consider the element $a^{-1} n a$, where $a \in G$, $n \in N$. Then, if it is in $N$, we are done. Otherwise, it is in $Na$, so $a^{-1}na = ma$ for some $m \in N$. Cancel out the $a$, and take the $n$ to the other side to get $a^{-1} = mn^{-1}$. Since $m,n \in N$, it follows from this that $a \in N$, but then obviously $a^{-1}na \in N$. This completes the proof of part $b$ with your observation. – 2017-02-21
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    Then basically the last part arrives at a contradiction since a is not in N? – 2017-02-21
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    @Sam Ah, yes, right, that can be put down as a contradiction. – 2017-02-21
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    okay I will just have to look over the proof again and make sure I read what you wrote out in the comments to make sure I understand the proof completely. Thank you very much for the help! – 2017-02-21
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    You are welcome, my friend. – 2017-02-21
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    I was just looking over the proof again and for part 1 do you mean x inverse is in N? Also, could part a be even more condensed possibly by just showing it one way using the index 2? – 2017-02-21
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    If $x^{-1} \in N$, certainly $x \in N$ as $N$ is a group in it's own right. Yes, part a can be condensed by just showing it one way. I elaborated to ensure that you do not miss any crucial details. – 2017-02-22