Given any h in H, k in K Since H is a normal subgraph of G (k^-1)hk in H => (k^-1)hk(h^-1) in H -- (1) Since K is a normal subgraph of G hk(h^-1) in K => (k^-1)hk(h^-1) in K -- (2) From (1) and (2), we have (k^-1)hk(h^-1) in H ^ K Since H ^ K = {e} => (k^-1)hk(h^-1) = e => hk = kh
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Given any h in H, k in K
Since H is a normal subgraph of G
(k^-1)hk in H
=> (k^-1)hk(h^-1) in H -- (1)
Since K is a normal subgraph of G
hk(h^-1) in K
=> (k^-1)hk(h^-1) in K -- (2)
From (1) and (2), we have
(k^-1)hk(h^-1) in H ^ K
Since H ^ K = {e}
=> (k^-1)hk(h^-1) = e
=> hk = kh
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