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Question:

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

Discussion:

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 $$|HK|=48$$.