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About LinkBacksLet N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.
I'm a confused about NH, what kind of element belong to this set?
The set $\displaystyle NH = \{nh| n\in N \mbox{ and }h\in H\}$. If $\displaystyle N,H$ are subgroups of $\displaystyle G$ then not necessarily does it means $\displaystyle NH$ is a subgroup. If however, $\displaystyle N$ is normal, i.e. $\displaystyle gng^{-1} \in N$ for any $\displaystyle n\in N$ and $\displaystyle g\in G$ then $\displaystyle NH$ is a group.Originally Posted by tttcomrader
However, it turns out that $\displaystyle N$ need not be normal in $\displaystyle G$. It can have a weaker property but yet $\displaystyle NH$.
Theorem: If $\displaystyle N$ has the property that $\displaystyle hnh^{-1} \in N$ for any $\displaystyle n\in N$ and $\displaystyle h\in H$ then $\displaystyle NH$ is a group.
(Do you see why this would prove you case? Meaning, if $\displaystyle N$ is normal then certainly $\displaystyle hnh^{-1}\in N$. So of course it works).
Proof: All I will show is closure of $\displaystyle NH$ which is the only part which takes a little work. Let $\displaystyle x_1 \in NH$ and $\displaystyle x_2\in NH$ we want to show $\displaystyle x_1x_2\in NH$. But this means $\displaystyle x_1 = n_1h_1$ and $\displaystyle x_2 = n_2h_2$ for $\displaystyle n_i \in N$ and $\displaystyle h_i \in H$. Thus, $\displaystyle x_1x_2=n_1h_1n_2h_2$. But is this product in $\displaystyle NH$? Not necessarily, so we need to use the additional condition in the theorem. Which says $\displaystyle h_1n_2h_1^{-1}\in N$, i.e. $\displaystyle h_1n_2h_1^{-1} = n_3\implies h_1n_2 = n_3h_1$ for some $\displaystyle n_3\in N$. Thus, $\displaystyle n_1h_1n_2h_2 = n_1(h_1n_2)h_2 = n_1(n_3h_1)h_2 = (n_1n_3)(h_1h_2)$ which certainly is in $\displaystyle NH$. Q.E.D.
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