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If $A$ is a normal subgroup of $B$ then is it required for $A$ and $B$ to be groups under the binary operation multiplication? what if they are just groups under the binary operation addition, can there still exist normal subgroup?

Like, set of Rational numbers is a group under addition only and not multiplication.

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  • $\begingroup$ Could you explain what you mean by "groups with multiplication" and "groups with addition"? $\endgroup$ – Zev Chonoles Jul 25 '13 at 5:36
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    $\begingroup$ It doesn't matter what the binary operation of a group is called. Though generally we only call an operation "addition" if at least it's commutative, and in an abelian group all subgroups are normal. $\endgroup$ – anon Jul 25 '13 at 5:36
  • $\begingroup$ @ZevChonoles i hope i am clear now. $\endgroup$ – aarbee Jul 25 '13 at 5:40
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When you say a group, it always refers to a unique " group structure" on a said set having chosen a specific binary operation. The same set may have other group structures borne of other binary operations. It is only when we talk about rings (or more so, fields) that we have to worry as to which binary operation "addition" or "multiplication", we are talking about. But that is not the point here.

When you are talking about a subgroup N being normal to a group G, it just means that $ \forall g \in G, \forall n\in N \implies gng^{-1} \in N $. The group operation there is the same as that which you invoked while you defined the group structure on your set.You can also look at normalcy as requiring every right coset to be a left coset(if you are aware of that term). Normalcy is a nice property as it facilitates construction of quotient groups, which are important in characterizing groups. I hope this helps.

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  • $\begingroup$ actually i also looked at the same definition. and gng(-1) made me think as if it is only for multiplication. but thanks to your explanation, i can now see that it is true for addition as well. $\endgroup$ – aarbee Jul 25 '13 at 6:08
  • $\begingroup$ btw @Vishesh, can u confirm me if (Rationals)/(Inegers) is a quotient group? that is to say it has elements like Z+a types. where Z is integer set and 'a' belongs to Rational no. set. $\endgroup$ – aarbee Jul 25 '13 at 6:19
  • $\begingroup$ Well I think the other answer has told you that. Anyways as the additive group of integers is a subgroup of the Additive group of rationals which is Abelian, the former is normal in the latter and hence the quotient group you asked about indeed exists. $\endgroup$ – Vishesh Jul 25 '13 at 7:04
  • $\begingroup$ in other answer, i was discussing normalcy, here i was asking about quotient group. and i am thankful that u cleared my doubt. $\endgroup$ – aarbee Jul 25 '13 at 7:05
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The short answer is that if A is an Abelian group, then every subgroup B of A is normal in A. When you see two elements a*b , or just ab , with * the group operation, then * can be anything (anything that satisfies the group axioms.)

Actually, if * is the group operation, then notice that B being normal in A is equivalent to saying that:

$ a*b*(a^{-1})$ is in B, for all a in A, for all b in B .

Now, let b be any element of B, and assume the group is Abelian. Then, using:

$ a*b*(a^{-1})$, we can rewrite , using commutativity ("Äbelianness" )

$ a*b*(a^{-1})$=$ a*(a^{-1})* b$ , which is, tautologically, in B

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  • $\begingroup$ so u think we can say set of integers is a normal subgroup of set of rational numbers? $\endgroup$ – aarbee Jul 25 '13 at 5:51
  • $\begingroup$ Yes, using addition as group operation. But it would be a nice exercise to first check that the integers are an actual subgroup of the rationals. $\endgroup$ – user87588 Jul 25 '13 at 5:57
  • $\begingroup$ oh yes, they are. $\endgroup$ – aarbee Jul 25 '13 at 6:10
  • $\begingroup$ Yes, I agree; I just thought it is a good idea, helpful, to prove that it is a subgroup. $\endgroup$ – user87588 Jul 25 '13 at 6:11

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