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Question:

Let $N\triangleleft G$ and $K\triangleleft G$. If $N\cap K=\langle e\rangle$ and $N\vee K=G$, then $G/N\cong K$.

I used the second theorem of isomorphisms and this one (From Hungerford's book):

Thus:

$K\cong K/\langle e\rangle\cong G/N$ (by second theorem of isomorphisms)

So I only use $N\triangleleft G$, am I wrong?

thanks in advance

Ok, checked it: it's just the subgroup generated by them...which is the product if, for example, at least one of the groups is normal. — DonAntonio, Apr 21 '14 at 20:21

score 4 · Accepted Answer · answered Apr 21 '14 at 20:16

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You know that $G=NK$ because $N$ is normal in $G$, so $N\vee K=NK$; by the homomorphism theorem $$ \frac{G}{N}=\frac{NK}{N}\cong\frac{K}{K\cap N}=\frac{K}{\{e\}}\cong K. $$ Indeed, the normality of $K$ is not necessary.

answered Apr 21 '14 at 20:16

egreg

238,574

In fact $K$ is said to be complement of $N$ in $G$. Only the normality of $N$ is required. – Lorban Apr 21 '14 at 20:25

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