If sum of order of two subgroup of g is greater than the order of g
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Certainly the set HKHK has |H||K||H||K|symbols. However,not all symbols need represent distinct group elements. That is, we may have hk=h′k′hk=h′k′ although h≠h′h≠h′ and k≠k′k≠k′. We must determine the extent to which this happens.
For every t∈H∩Kt∈H∩K, hk=(ht)(t−1k)hk=(ht)(t−1k), so each group element in HKHK is represented by at least |H∩K||H∩K| products in HKHK.
But hk=h′k′hk=h′k′ implies t=h−1h′=k(k′)−1∈H∩Kt=h−1h′=k(k′)−1∈H∩K so that h′=hth′=ht and k′=t−1kk′=t−1k. Thus each element in HKHK is represented by exactly |H∩K||H∩K| products. So,
|HK|=|H||K||H∩K|.|HK|=|H||K||H∩K|.
If we have hk=h′k′hk=h′k′ and we multiply this by h−1h−1 from left and by k′−1k′−1 from right, we get
kk′−1=h−1h.kk′−1=h−1h.
Maybe it should be stressed that t∈Ht∈H, since t=h−1h′t=h−1h′; and t∈Kt∈K since t=kk′−1t=kk′−1. (Which means t∈H∩Kt∈H∩K
SHAKUNTALAM GHOSH.
For every t∈H∩Kt∈H∩K, hk=(ht)(t−1k)hk=(ht)(t−1k), so each group element in HKHK is represented by at least |H∩K||H∩K| products in HKHK.
But hk=h′k′hk=h′k′ implies t=h−1h′=k(k′)−1∈H∩Kt=h−1h′=k(k′)−1∈H∩K so that h′=hth′=ht and k′=t−1kk′=t−1k. Thus each element in HKHK is represented by exactly |H∩K||H∩K| products. So,
|HK|=|H||K||H∩K|.|HK|=|H||K||H∩K|.
If we have hk=h′k′hk=h′k′ and we multiply this by h−1h−1 from left and by k′−1k′−1 from right, we get
kk′−1=h−1h.kk′−1=h−1h.
Maybe it should be stressed that t∈Ht∈H, since t=h−1h′t=h−1h′; and t∈Kt∈K since t=kk′−1t=kk′−1. (Which means t∈H∩Kt∈H∩K
SHAKUNTALAM GHOSH.
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