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# Let H and K be subgroups of a group G- If G-HK and g-hk- where h is in.docx ×

# Let H and K be subgroups of a group G- If G-HK and g-hk- where h is in.docx

Let H and K be subgroups of a group G. If G=HK and g=hk, where h is in H and k is in K, is there any relationship among |g|,|h|,and |k|? What if G=HxK?
Solution
If hk = h\'k\' then (h\')^-1h = k\'(k)^-1 = x. Since x is in (H intersect K) = 1, then h = h\', k = k\'. This means that any element g in G can be expressed uniquely as g = hk with h in H an k in K.

Since both H and K are normal then (hk)(h\'k\') = h(kh\'k^-1)kk\' where kh\'k^-1 is in H. Using the same reasoning (hk)(h\'k\') = hh\'(h\'^-1kh\')k\' where h\'^-1kh\' is in K.

Since h(kh\'k^-1)kk\' = hh\'(h\'^-1kh\')k\', and the expression for (hk)(h\'k\') as h\"k\" is unique, then looking at the factors in each expression it follows that kh\'k^-1 = h\' and h\'^-1kh\'= k.

So G = H?K.
.

Let H and K be subgroups of a group G. If G=HK and g=hk, where h is in H and k is in K, is there any relationship among |g|,|h|,and |k|? What if G=HxK?
Solution
If hk = h\'k\' then (h\')^-1h = k\'(k)^-1 = x. Since x is in (H intersect K) = 1, then h = h\', k = k\'. This means that any element g in G can be expressed uniquely as g = hk with h in H an k in K.

Since both H and K are normal then (hk)(h\'k\') = h(kh\'k^-1)kk\' where kh\'k^-1 is in H. Using the same reasoning (hk)(h\'k\') = hh\'(h\'^-1kh\')k\' where h\'^-1kh\' is in K.

Since h(kh\'k^-1)kk\' = hh\'(h\'^-1kh\')k\', and the expression for (hk)(h\'k\') as h\"k\" is unique, then looking at the factors in each expression it follows that kh\'k^-1 = h\' and h\'^-1kh\'= k.

So G = H?K.
.