I want to classify the groups of order 42, where the 7-Sylow subgroup $P_7$ is normal, the 2-Sylow subgroup is not normal, and we have 7 3-Sylow subgroups $P_3$.
This is what I did:
Since $P_7$ is normal, $M = P_7P_3$ is a subgroup. Since $|P_7P_3|=21$, it must be normal in G. We also know that $P_2$ is a subgroup of roder G and $P_2 \cap P_7P_3 = \{e\}$, because otherwise $P_2$ would be a subgroup of $P_7P_3$ but 2 doesn't divide 21. So $G \cong M \rtimes_{\alpha} P_2$.
Since M has order 21, we need to classify groups of order 21 to look at the cases for different possibilites of M.
|M|= 21 = (3)(7)
$n_7 \equiv 1 \pmod 7$ and $n_7|2 \implies n_7=1$.
$n_3 \equiv 1 \pmod 3$ and $n_3|7 \implies n_3=1,7$
Case a:
$n_3 = 1 \implies M \cong Z_3 \times Z_7 \cong Z_{21}$
Case b:
Since $P_7 \vartriangleleft M$, and $P_7 \cap P_3 = \{e\}$ (since 3 doesn't divide 7), $M \cong Z_7 \rtimes_{\bar{alpha}} Z_3$.
If $Ker \bar{\alpha} = Z_3$, then $M \cong Z_{21}$.
If $Ker \bar{\alpha} = \{e\}$, then $\bar{\alpha}: Z_3 \rightarrow Z^{\times}_7$ is one-to-one.
Case 1:
$M \cong Z_{21}$, so $G \cong Z_{21} \rtimes_{\alpha} Z_2$.
If $ker \alpha = Z_2$, then $G \cong Z_{42}$.
If $ker \alpha = \{e\}$, then $\alpha: Z_2 \rightarrow Z^{\times}_{21}$ is one-to-one.
Case 2:
$M \cong Z_7 \rtimes_{\alpha} Z_3$.
$G \cong (Z_7 \rtimes_{\bar{\alpha}} Z_3) \rtimes_{\alpha} Z_2$.
But I'm not sure how to simplify this…In other words, I want to simplify this to know if it is isomorphic $Z_7 \rtimes_{\alpha} Z_6$…
Thanks in advance
Best Answer
You're off to a good start. Here's how it goes.
$M$ is isomorphic to either $\mathbb Z_{21}$ or the Frobenius group, $F_{21}=\left<a,b\mid a^7=b^3=e,ba=a^2b\right>$. And clearly, $P_2\cong\mathbb Z_2$, say, $P_2=\left<c\right>$.
If $M\cong\mathbb Z_{21}$, say $M=\left<a\right>$, then $cac^{-1}=a^t$ for some $0\le t<21$ since $M$ is normal, thus $ca=a^tc$. Since $a=c^2a=ca^tc=a^{t^2}cc=a^{t^2}$, $t^2\equiv 1\pmod{21}$ so that $t$ is either $1$, $8$, $13$ or $20$. These four cases yield the groups $\mathbb Z_{42}$, $\mathbb Z_7\times S_3$, $D_7\times\mathbb Z_3$, and $D_{21}$ respectively.
Now suppose $M\cong F_{21}$. Then the map $\sigma:F_{21}\to F_{21}$ sending $x\to cxc^{-1}$ is an automorphism and $\sigma^2$ is the identity map $\iota$.
Now, $\sigma(a)=a^t$ for some $1\le t\le 6$ (because those are the only elements of $F_{21}$ of order $7$) and $\sigma^2=\iota$ implies $t^2\equiv 1\pmod 7$ so that $t$ is either $1$ or $6$. If $t=1$, then $\sigma(b)=a^ib$ for some $i$ (it can be shown that $\sigma(b)a=a^2\sigma(b)$ which would fail if $\sigma(b)=a^ib^2$); $\sigma^2=\iota$ implies $2i\equiv 0\pmod 7$, so that $\sigma(b)=b$ and $\sigma=\iota$. In this case, $M$ and $P_2$ commute, and $G\cong F_{21}\times\mathbb Z_2$.
Now suppose $t=6$. Then it doesn't matter what $\sigma(b)$ is; $\sigma^2=\iota$ would follow anyway, because if $\sigma(b)=a^ib$, then $\sigma(a^ib)=\sigma(a)^i\sigma(b)=(a^6)^ia^ib=a^{7i}b=b$. Hence, $\sigma^2(a)=a$ and $\sigma^2(b)=b$, so that $\sigma^2=\iota$. It can be shown that there are $7$ possibilities for $\sigma$ and each yields an isomorphic result for $G$, so suppose $\sigma(b)=b$. Then $bc=cb$, $\left<bc\right>$ is a cyclic subgroup of order $6$, $bca=a^5bc$ and $G$ is isomorphic to the holomorph $\mathbb Z_7\rtimes\mathbb Z_6$.
Summarizing, the six groups of order $42$ are $\mathbb Z_{42}$, $\mathbb Z_7\times S_3$, $D_7\times\mathbb Z_3$, $D_{21}$, $F_{21}\times\mathbb Z_2$ and the holomorph $\mathbb Z_7\rtimes\mathbb Z_6$. It can be shown quite feasibly that no two of those are isomorphic.