Internal semidirect product and roots of unity

[mahler1]

Homework Statement

Let $G=G_{12}$, $H_1=G_3$, $H_2=G_2$. Decide if there are groups $K_1$, $K_2$ such that $G$ can be expressed as the internal semidirect product of $H_i$ and $K_i$.

The Attempt at a Solution

Suppose I can express $G_{12}$ as an internal semidirect product between $G_3$ and $K$, with $K$ subgroups. Then it must be $G_3 \cap K={1}$ and $G_12=G_3.K$. I know that for $n,m$, $G_{n} \cap G_{m}={1}$ if and only if $(n:m)=1$. Taking this condition into account and the fact that $K$ has to be a subgroup of $G_{12}$, the candidates for $K$ would be $G_i$ with $i \in \{2,4\}$. I am not so sure how to realize if one of these subgroups could work. The case $H_2=G_2$ is analogous, I would appreciate suggestions.

 

[mighty2000]

To determine if $G$ can be expressed as an internal semidirect product of $H_i$ and $K_i$, we need to check if the conditions for an internal semidirect product are satisfied. These conditions are:

1. $H_i$ and $K_i$ are subgroups of $G$.
2. $G=H_iK_i$.
3. $H_i \cap K_i = \{1\}$.
4. $H_i$ and $K_i$ normalize each other, i.e. $K_i$ is a normal subgroup of $H_iK_i$.

Based on the given information, we can see that $H_i$ and $K_i$ are indeed subgroups of $G$, as $G_{12}=G_3K_1=G_2K_2$. However, we cannot say for certain if $G=H_iK_i$, as we do not have enough information about the structure of $G$. Similarly, we cannot determine if $H_i \cap K_i = \{1\}$ without knowing more about the subgroups $H_i$ and $K_i$.

In order to determine if $G$ can be expressed as an internal semidirect product of $H_i$ and $K_i$, we need to investigate the structure of $G$ further. We can use the given information about the subgroups $H_i$ and $K_i$ to narrow down the possibilities for $G$, and then check if the conditions for an internal semidirect product are satisfied for those possibilities.