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I am currently trying to understand lattice theorem (fourth isomorphism theorem), states that if N is a normal subgroup of a group G, then there exists a bijection from the set of all subgroups A of G such that A contains N, onto the set of all subgroups of the quotient group G/N. I was trying on this example:

Suppose $G=C_9 \times C_{81}$ and $N=C_9 \times C_{27}$. Then $G/N$ is isomorphic to $C_3$, having 1 proper subgroup. Thus $G$ has 1 proper subgroup that contains $N$, which is true.

However, if we let $N'=C_3 \times C_{81}$, we have $G/N'$ to be isomorphic to $C_3$ as well, leading to $G$ has only 1 proper subgroup that contains $N'$. In fact, $G$ has 3 proper subgroups that contain $N'$.

Could anyone help me out? Many thanks.

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In your second example, there are still no proper subgroups of $C_9\times C_{81}$ which contain $N'=C_{3} \times C_{81}$. To see this, suppose that $N' \subset H \subsetneq C_9 \times C_{81}$. Then, by Lagrange's theorem:

$$3=\frac{|G|}{|N'|}= \frac{|G|}{|H|}\cdot \frac{|H|}{|N'|}.$$

Note that the first factor on the right side is at least $2$, by the assumption that $H$ is proper. Since $3$ is prime, this forces the second factor to be $1$. Hence $$\frac{|H|}{|N'|}=|H/N'|=1,$$ and $H = N'=C_3 \times C_{81}$.

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  • $\begingroup$ Thanks for the reply.For the first example, I could make the bijection by: 1. e -> N 2. $C_3$ -> G $\endgroup$ Aug 25 '14 at 16:32
  • $\begingroup$ @ChunYongChew yes, that's exactly the bijection. $\endgroup$ Aug 25 '14 at 16:35
  • $\begingroup$ Hi Morgan, But for the second example, I computed this group in GAP: gap> x:=DirectProduct(c9,c81); <pc group of size 729 with 6 generators> the subgroups that are isomorphic to N': Group( [ f6, f5, f4, f1*f3, f2 ] ) Group( [ f6, f5, f4, f1*f3^2, f2 ] ) Group( [ f6, f5, f4, f2, f3 ] ) G has 3 different subgroups that is isomorphic to N'. How is the bijection being defined? $\endgroup$ Aug 25 '14 at 16:41
  • $\begingroup$ @ChunYongChew there may be other subgroups that are isomoprhic to $N'$, but none except $N'$ can contain $N'$. Note that the isomorphism theorem gives a bijection between subgroups of $G/N'$ and subgroups of $G$ which contain $N'$. The bijection is just like in your first example: 1. $e \mapsto N'$ and 2. $C_3 \mapsto G$. $\endgroup$ Aug 25 '14 at 16:46
  • $\begingroup$ Thanks for the explanation. $\endgroup$ Aug 25 '14 at 17:05

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