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Learn MoreLet's discuss a Common but deadly question in Group theory.

**Question**: **Is it possible to get an infinite group which has elements of finite order? **

Particularly for this problem I am going to take two very fundamental groups

1) (Q , +) (Rationals under addition)

2) (Z , +) (Integers under addition)

Now can you prove that (2) is a normal subgroup of (1). (Hint: Use the definition of normal subgroups)

So **Q/Z** under the binary operation will be a quotient group.

Now a little of topic discussion, if you know what will be the elements then you sure know the identity element of Q/Z. DON'T MAKE MISTAKE THAT '0' IS THE IDENTITY OF Q/Z. THE IDENTITY IS Z ITSELF AS THE ELEMENTS OF Q/Z ARE ALL IN FORM OF SETS.

For an example let p/q our usual rational number. Then the elements of Q/Z will be of the form (p/q) + Z

So as I have told you before that the identity is Z so for fun add (p/q) + Z two times

It gives (2p/q) + Z isn't so. Then what will happen if you add it q times? It will be p + Z. Now p is an integer itself. Then p + Z is Z itself. So you get an arbitrary element in Q/Z which has finite order q.**(As q is the smallest integer to do so because gcd (p,q) is 1).** Hope all of new comers in group theory will understand this.

This kind of question is very **important for JAM, TIFR mainly.**

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