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Cardinality of product of subgroups (TIFR 2014 problem 14)


Let \(G\) be a group and \(H,K\) be two subgroups of \(G\). If both \(H\) and \(K\) has 12 elements, then which of the following numbers cannot be the cardinality of the set \(HK=\{hk|h\in H , k\in K\}\)

A. 72

B. 60

C. 48

D. 36


We have \(|H|=|K|=12\).

We know that \(|HK|=\frac{|H||K|}{|H\cap K|}\).\(…*\)

Or, in other words \(|HK||H\cap K|=|H||K|\).

So, at-least we expect to have \(|HK|\) divides \(|H||K|=12^2=144\).

Here, \(72,48,36\) all divide \(144\) but \(60\) does not divide \(144\) therefore \(|HK|\) can not be \(60\).

Now, the question still remains whether there exists subgroups which give rise to \(|HK|=72,48,36\). The answer is yes they do exist. And this is in fact given by the formula \(*\) above. All we need to do is take two subgroups which have only \(\frac{144}{72},\frac{144}{48},\frac{144}{36}\) elements common respectively.

For example take \(H=D_{2.6}\) and \(K=\{1,s\}\times\mathbb{Z/6Z}\) where \(s\) is the reflection (element of order 2) and we then get example of \(|HK|=72\). Here we considered \(D_{12}\) as \(D_{12}\times\{\bar{0}\}\). The intersection is \(\{1,s\}\times\{\bar{0}\}\) which has cardinality 2.

Take \(H=A_4\) and \(K=\{(1),(12)(34),(13)(24),(14)(23)\}\times \mathbb{Z/3Z}\). Then we get example of \(|HK|=36\). Here we considered \(A_4\) as \(A_4\times\{\bar{0}\}\). The intersection is \(\{(1),(12)(34),(13)(24),(14)(23)\}\times \{\bar{0}\}\) which has cardinality 4.

For the same \(H\) taking \(K=\{(1),(123),(132)\}\times \mathbb{Z/4Z}\) we get \(|HK|=48\).

October 27, 2017

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