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The subgroup correspondence theorem says that for a group $G$ and a normal subgroup $N\subset G$, the canonical projection $\pi\colon G\rightarrow G/N$ establishes a bijection between the subsets of subgroups of $G$ containing $N$ and the subgroups of $G/N$ (mapping a subgroup $U$ of $G$ containing $N$ to $\pi(U)$ and a subgroup $V$ of $G/N$ to $\pi^{-1}(V)$), and that this bijection identifies normal subgroups on both sides.

Now I'm searching for simple, yet pedagogically valuable examples to illustrate this, on the level of an introductory course on algebra. In particular, in view of the addendum for normal subgroups, the groups involved should be non-abelian.

A first example I found is to use the symmetric group $S_4$ as $G$ and let $N:=V_4$ be the Klein four group, so that $G/N\simeq S_3$. Then, one may identify $A_3\lhd S_3$ with $A_4\lhd S_4$ and the three subgoups of order $2$ in $S_3$ with the three dihedral groups $D_4$ in $S_4$.

However, I would really like to know about other and possibly even shorter examples! So my question is: What are your favorite examples for this? Many thanks in advance!

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The smallest possible example (where $G/N$ is not cyclic) is the quaternion group $G=Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$ and $N=\{\pm 1\}$. Then $G/N\cong C_2\times C_2$, the three proper non-trivial subgroups of which correspond to the subgroups $\langle i\rangle$, $\langle j\rangle$, $\langle k\rangle$ in $G$.

I think this is a particularly nice example because as a corollary it proves by counter-example that the following often falsely believed statement is false:

"If $G$ is a group such that every subgroup is normal then $G$ is abelian."

It's easily seen that $-1$ is in any non-trivial subgroup of $G$, so every non-trivial subgroup contains $N$, and therefore corresponds to a subgroup of $G/N$ which is abelian, hence every subgroup of $G$ is normal even though $G$ is non-abelian.

Addition

Another nice relatively simple use of this result is to find all normal subgroups of the dihedral group $D_{2^{n+1}}$ of order $2^{n+1}$.

Using $G=D_{2^{n+1}}=\langle a,b|a^{2^n}=b^2=1,b^{-1}ab=a^{-1}\rangle$, suppose $1\ne N\trianglelefteq G$. If $a^ib\in N$ for some $i$ then $N\ni (a^ib)^{-1}a(a^ib)a^{-1}=a^{-2}$ so $a^{2^{n-1}}\in N$. That is any non-trivial normal subgroup of $G$ contains $M=\langle a^{2^{n-1}}\rangle$ so corresponds to a normal subgroup of $G/M\cong D_{2^n}$. Proceeding by induction you can check that the proper non-trival normal subgroups of $D_{2^{n+1}}$ are precisely the subgroups of the form $\langle a^i\rangle$ for some $i$ and $\langle a^2,b\rangle$.

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  • $\begingroup$ Very nice examples, thanks a lot! $\endgroup$ – user103697 May 1 '18 at 14:06

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