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Can there exist normal subgroup having index greater than two.... It's just a question that occurred in my head while doing problems , I need no proof as such.

VCA
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    There is no need to downvote this question or ask to close it! While it is a very (very) naive question it is a valid question. Given the close votes this convinces me that people do abuse the system. – abnry Dec 30 '15 at 15:34
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    Protip: when a question "just occurs to you" the next step should be checking it against examples you know. Since you probably know what the integers are, you should have immediately seen an answer to your question. – rschwieb Dec 30 '15 at 16:23
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    @nayrb The reason for voting to close is not because the question is naive but because it is lacking in context, which makes it difficult to provide the a helpful answer. Surely, with any question of the form "is the following true...?" at any level, you should include some indication (no matter how naive) of why you think it might be true? – Derek Holt Dec 30 '15 at 16:25
  • @DerekHolt I don't see why more "context" is needed. The OP is doing homework problems and clearly all the examples there involve normal groups of index two. All that's needed to answer the question is give some examples of when this doesn't occur. – abnry Dec 30 '15 at 16:35
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    @nayrb To be fair, I kind of doubt the OP is using a reference that has a lengthy list of normal subgroups, all of index $2$. More likely they have one or two examples, or maybe they're most familiar with the infinite class of examples $A_n \subset S_n$ of the alternating groups inside the symmetric groups, and they didn't really think much more about it. It is true that some added context could potentially make it possible to give a more pedagogically useful answer; I just don't think it's worth asking that of this question. Just give some counterexamples for OP to chew on and move on. – Dustan Levenstein Dec 30 '15 at 17:44
  • I understand that this was not a good question to ask , I realised just now the stupidity in asking something like this when there were other questions in my text like to prove that the particular normal subgroup for index 3 and 4 . – VCA Jan 02 '16 at 04:45

5 Answers5

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Yes, it is possible. For abelian groups you can take any subgroup of index greater than two. It is normal. There are also many examples for non-abelian groups. The symmetric group $S_4$ has many normal subgroups of index greater than $2$, see here.

Community
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Dietrich Burde
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An even simpler answer: the trivial subgroup is always a normal subgroup. So your question is equivalent to the question "Does there exist a group of order greater than 2?"

Of course the answer is yes. Consider $G = \mathbb Z/3\mathbb Z$.

Dustan Levenstein
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Let $G=H \times C_n$, where $H$ is any group and $C_n$ is the cyclic group of order $n$. Then $H$ is normal and $|G:H|=n$.

Nicky Hekster
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You might recall that the index of a normal subgroup $N < G$ equals the order of the quotient group $G/N$. Also, for any surjective homomorphism $G \to Q$ with kernel $N$, the group $Q$ is isomorphic to $G/N$, and so the index of $N$ equals the order of $Q$.

So your question is equivalent to asking:

  • Can there exist surjective homomorphisms $G \to Q$ such that $Q$ has order greater than $2$?

Yes there can. For example, the map $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $f(x,y)=y$ is a surjective homomorphism, with respect to the group operations of vector addition on $\mathbb{R}^2$ and ordinary addition on $\mathbb{R}$. The kernel of $f$ is the $x$-axis, and its index equals the order of the image group $\mathbb{R}$, which is uncountably infinite.

Lee Mosher
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You're probably confused by the fact that an index $2$ subgroup is automatically normal; however this is the exception, rather than the norm.

A famous theorem by Feit and Thompson implies that every group with odd order has a proper and non trivial normal subgroup and, of course, such a subgroup cannot have index $2$.

More easily, an abelian group has subgroup of every possible order (taking Lagrange's theorem into account, of course):

If $G$ is a finite abelian group of order $n$ and $d$ is a divisor of $n$, then $G$ has a subgroup $H$ of order $d$ (and index $n/d$).

If $G$ is cyclic, the proof is very simple: take a generator $x$ of $G$; then the subgroup $\langle x^{n/d}\rangle$ generated by $x^{n/d}$ has order $d$ (and index $n/d$).

egreg
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