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Can you give an example of two subgroups $H$ and $K$ where $o(H \cap K) > 1$ and hence find the order of $HK$?

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Please do not use all-caps titles. –  Did Nov 25 '12 at 15:48
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DID YOU KNOW THAT ALL CAPS IS CONSIDERED SHOUTING ON THE INTERNET?? –  Asaf Karagila Nov 25 '12 at 15:50
    
What, Asaf? I didn't quite "hear" what you said... ;-) –  amWhy Nov 25 '12 at 15:52
    
@AsafKaragila - I found that the hard way –  Belgi Nov 25 '12 at 16:37

1 Answer 1

up vote 3 down vote accepted

Your question is ambiguous:

$(1)$ Are $H$ and $K$ both subgroups of the same group $G$? If so, Are you given any information about $G$?

$(2)$ And what do you mean by $HK$? Assuming you mean $HK$ is a subgroup of the same group of which $H$ and $K$ are subgroups, then perhaps you mean for $HK = \{h*k | h\in H \;\text{and}\;k\in K\}$?

$(3)$ Also "O(H and K)" is ambiguous. Do you mean that the $\text{ord}(H) > 1$ and $\text{ord}(K) > 1$? Or do you mean $\text{ord}(H \cap K) > 1$?

CLARIFIED "O(H and K) = $\text{ord}(H \cap K)$...


That said:

First, assuming $H \le G$ and $K \le G$ ($\le$ meaning "is a subgroup of"), then $H \cap K \le G$. But that intersection may be only the identity of $G$, and hence of order 1. $|H \cap K| = \text{ord}(H\cap K) > 1$ if and only if $\text{gcd}(|H|, |K|) \neq 1$.

We can show that if $H$ and $K$ are subgroups of an abelian group $(G, *)$, and if $HK$ denotes a subgroup, then $$HK = J = \{h*k | h\in H \;\text{and}\;k\in K\} \le G.$$ So if the order of $H = a >1$, and the order of $K = b > 1$, then the order of $HK = J = \text{lcm}(a, b)$.

Can you think of a finite abelian group $G$ with subgroups $H, K$ such that if you know $|H| = a > 1$ and $|K| = b > 1$, then $|HK|>1$?

Take, for example, $G = \mathbb{Z}_{24}, \;H = \langle 4\rangle = \{0, 4, 8, 12, 16, 20\} \; K = \langle 6 \rangle = \{0, 6, 12, 18\},\; H\cap K = \{0, 12\}$.

Clearly, $H \le G, \; K \le G$. $|H| = 6>1, \; |K| = 4>1, \; |H\cap K| = 2 > 1$.

Then $HK = \{h+k|h \in H, \; k \in K\} = \langle 10 \rangle = \langle 2 \rangle = \{0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22\}$ and $|\langle 2 \rangle | = 12 = \text{lcm(6, 4)} > 1$.


Note: you can also take for an example any $H, K \le \mathbb{Z}$.

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ya..by (H and K), i meant intersection only and i didn't get how o(HK) is lcm(a,b) it should be [O(H).O(K)]/O(H and K) right?? –  ANKITA Nov 25 '12 at 20:39
    
Yes, if you mean (H and K) = $H \cap K$, then yes, $[O(H) \cdot O(K)]/O(H\cap K)]$ which in my example, would be $O(HK) = (6\cdot4)/2) = 12$, since $H \cap K$ is a subgroup of G when H and K are subgroups of G and O$(H\cap K)$ = gcd(O(H), O(K)) –  amWhy Nov 25 '12 at 20:54
    
Just to be clear, in the example I give above, $\text{ord}(H \cap K) = \text{gcd}(\text{ord}(H), \text{ord}(K)) = \text{gcd}(6, 4) = 2$. –  amWhy Nov 25 '12 at 21:09
    
sorry to ask again and again, but how is O(H and K) = 2? –  ANKITA Nov 25 '12 at 21:14
    
$H \cap K = \{0, 12\} \le \mathbb{Z}_{24}$, while $HK = \{0, 2, 4, ... , 20, 22\}\le \mathbb{Z}_{24}$. $H\cap K\ne HK$, $O(H\cap K)=2, O(HK) = 12$. –  amWhy Nov 25 '12 at 21:24

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