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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).


  • What is this topic: Abstract Algebra
  • What are some of the associated concept: Finite Order,Order of Subgroup
  • Book Suggestions: Contemporary Abstract Algebra by Joseph A. Gallian