How many non-isomorphic groups of order 122 are there?

Let $G$ be a group of order 122.No of Sylow 61 subgroups of order 61=1 and hence it is normal say it is $H$.

No. of Sylow 2 subgroups of order 2 is either 1 or 61.If it is 1 then $G\cong \mathbb Z_2\times \mathbb Z_{61}$

I have two questions frm now on:

Since we have two different choices for number of Sylow 2 subgroups can we conclude from here that there exist two non-isomorphic groups of order 122? Please help

My problem is when it is 61.Then we have 61 Sylow 2 subgroups of order 2.In this case how should I find $G$?Any help

  • $\begingroup$ Do you know any other groups of order 122? $\endgroup$ – Christopher May 20 '15 at 15:49
  • $\begingroup$ How about $D_{61}$ ? $\endgroup$ – user228113 May 20 '15 at 15:52
  • 2
    $\begingroup$ You can see, for example math.buffalo.edu/~badzioch/MTH619/Lecture_Notes_files/… . $\endgroup$ – Alex W May 20 '15 at 15:55
  • $\begingroup$ You showed that there is a normal subgroup of order 61 (a prime) and another subgroup of order 2. So $G$ is a semidirect product $Z_{61}\rtimes Z_2$. The possible semidirect products depend on the chosen homomorphisms $Z_2\to Aut(Z_{61})$. How many such homomorphisms do you know? $\endgroup$ – j.p. May 20 '15 at 15:59
  • $\begingroup$ There is nothing special about $61$. You might just as well consider groups of order $2p$ for an odd prime $p$. $\endgroup$ – Derek Holt May 20 '15 at 16:28

It can be shown that if the order of a group is $2p$ for some odd prime $p$, then either

$G \cong Z_{2p}$ or $G \cong D_p$

By the first Sylow Theorem, $G$ has subgroups $H, K$ such that:

$$\newcommand{\angle}[1]{\langle #1\rangle} H = \angle{a},\,K = \angle{b},\,|a| = p,\,|b| = 2$$

Since [G:H] = 2, H is a normal subgroup of G; this means

$$\newcommand{\set}[1]{\left\{ #1\right\}}HK = \set{hk \mid h \in H, k \in K} < G$$

Since every element of $H$ has order 1 or $p$, and every element of $K$ has order $1$ or $2$,

$$H \cap K = \set{g \in G \mid |g| = 1} = \set{e}$$

This gives us

$$|HK| = \frac{|H|\cdot|K|}{|H \cap K|} = 2p/1 = 2p = |G|,$$ so $HK = G$

$$G = \set{a^i b^j \mid |a| = p, |b| = 2, ba = ?}$$

It remains to be seen what $ba$ equals; that will give us the group table and the group.

Since $H$ is normal in $G$, $bab^{-1}\in H$, $bab^{-1} = a^i$ for some integer $i$

$$bab^{-1} = a^i, a = b(a^i)b^{-1}, a = b [b(a^i)b^{-1}]^i b^{-1}, a = a^{i^2}, a^{i^2 - 1} = e$$

This means that $|a| = p\mid i^2-1$, so $i =p+1\equiv 1\pmod{p}$ or $i =p -1$

$$i = 1,\quad ba = ab,\quad G\text{ Abelian}\implies G \cong \Bbb{Z}/2\Bbb{Z} \times \Bbb{Z}/p\Bbb{Z} \cong \Bbb{Z}/2p\Bbb{Z}$$ $$i = p-1,\quad ba = a^{p-1}b,\quad ba = a^{-1}b \implies G \cong D_p$$

Substitute 61 for $p$ and you're good to go


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