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I'm stuck with the following I hope someone could help me.

Let $N$ a normal subgroup of $G$. Show that if $[G:N]=4$, exists a normal subgroup $M$ of $G$ s.t. $[G:M]=2$. My idea: Since $G/N$ has orden 4, either is isomorphic to $\mathbb{Z}/4$ or $\mathbb{Z}/2 \times\mathbb{Z}/2$ and in any case we have a subgroup of order $2$ and for instance the same applies to $G/N$ but from here I'm not sure how to get the normal subgroup $M$ of $G$ which have exactly two cosets. any idea?

I think I got it. $G \to_\pi G/H\to _f\{\text{klein 4 group or cyclic group}\}$. Either if we have a cyclic group or the Klein 4 group (both commutatives) we can find a subgroup of order $2$, say $H$ (which obviously is normal), then $f^{pre}[H]=K\lhd G/H$, and define $M =\pi^{pre}[K]\lhd G$ (because the canonical map is a surjecion). To conclude we claim that $[G:M]= 2$. Since $4=[G:N]=[G:M][M:N]=[G:M]2$, we're done.

I think is correct, thanks

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    $\begingroup$ Use correspondence. Whenever $N\lhd G$ there is a bijective correspondence between the subgroups $H$ of $G/N$ and those subgroups $K$ of $G$ that contain $N$. More precisely the correspondence looks like $K\mapsto K/N=H$. For a full understanding (and proof) of this you may want to consider the projection $\pi:G\to G/N$, when $K=\pi^{-1}(H)$. Recall that if $f:G_1\to G_2$ is a surjective homomorphism of groups, and $N_2\lhd G_2$, then $N_1=f^{-1}(N_2)\lhd G_1$. $\endgroup$ – Jyrki Lahtonen Aug 26 '14 at 5:39
  • $\begingroup$ @JyrkiLahtonen do your refer to the fibers of the subgroup $H$ of order 2 of $G/N$, under the canonical projection. Am I right? $\endgroup$ – Jose Antonio Aug 26 '14 at 5:44
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    $\begingroup$ Possible duplicate with here $\endgroup$ – Golbez Aug 26 '14 at 5:45
  • $\begingroup$ I guess so? I would call it preimage, but it is the union of the preimages of elements of $M$, and those could be called fibers? $\endgroup$ – Jyrki Lahtonen Aug 26 '14 at 5:57
  • $\begingroup$ @JyrkiLahtonen, thanks :) $\endgroup$ – Jose Antonio Aug 26 '14 at 6:44
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Use the homomorphism theorems. By the correspondence theorem, if $N \trianglelefteq G$ and $\pi:G \rightarrow G/N$ is the canonical projection, then $\pi$ respects indices, factor groups, containment, and normality. Thus, if $H'$ is the image of $H$, then $H'$ is normal in $G':=G/N$ iff $H$ is normal in $G$, and the index $|G':H'|$ equals $|G:H|$. In your case $G/N \cong C_4$, which has a normal subgroup $H'$ of index 2. By the correspondence theorem, the (complete) preimage of $H'$ is a normal subgroup in $G$ and has index 2 in $G$.

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