Simple evidence that team of order $2p$ with nontrivial facility is cyclic?

We have a team $G$ of order $2p$, $p$ a weird prime, which has a nontrivial facility. We desire to show the team is cyclic.

This is a straightforward workout which can be promptly eliminated making use of hefty cannons. Nonetheless, given that I was inquired about it by pupils which hardly got to ratios, I question what straightforward means are to confirm this ; I could not locate any kind of.

Many thanks.

2019-12-02 02:50:24
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Answers: 3

Let $g$ be a nontrivial component of the facility. If $g$ creates the entire team, we are done ; or else $g$ creates a cyclic subgroup of order $2$ (resp. $p$). Therefore $G/\langle g \rangle$ is a cyclic subgroup of order $p$ (resp. $2$). It adheres to that $G$ is created by $g$ and also another component $h$ such that $h^p = g^k$ for some $k$ (resp. $h^2 = g^k$ for some $k$). Given that $g$ remains in the facility, $gh = hg$. Yet this indicates $h$ is additionally in the facility. Therefore $G$ is abelian.

If $h$ creates the entire team we are done. Or else $h$ has order $2$ (resp. $p$), given that it has order separating $2p$ and also can not have order $p$ (resp. $2$), and also $G \simeq C_2 \times C_p$ as wanted.

2019-12-03 04:21:23

Qiaochu is solution is less complex, yet the adhering to virtually identifies all teams of order $2p$, so I'll leave it below too.

Lemma 1 : The order of a component separates order of the team. No new evidence, yet very easy sufficient.

Lemma 2 : There exists a component of order $p$. Can prevent common evidence as adheres to : If there is a component of order $2p$ G is cyclic. If no component of order $p$ or $2p$ after that $a^2=e$ for all $a$, so $G$ is abelian ($abab=e$, so $ab=b^{-1}a^{-1}=ba$). After that either make use of the reality that you recognize all abelian teams, or the reality that after that $G$ is a vector room over the area $Z_2$, so has basis therefore has order $2^k$.

Call this component $a$.

Lemma 3 : Any subgroup of index 2 is regular. There are 2 cosets, and also among them is the subgroup itself. So appropriate and also left cosets synchronize.

Using this to the subgroup created by $a$, for $b$ not a power of $a$, we have $bab^{-1}=a^m$. The team has just $a^k$ and also $ba^l$'s, which are all distinctive. If $m=1$ they all commute and also $ba$ creates the cyclic team. If $m\neq1$ there is no facility. (The only action in category left is to show that various $m \neq 1$ bring about the very same team, which is not tough either.)

2019-12-03 04:21:11

Actually, this can be done really merely making use of the adhering to 2 realities, both of which adhere to from points the pupils need to recognize now (though they might not have actually seen a centralizer in the past, it is not also difficult to comprehend). First, any kind of subgroup of prime index is topmost. This adheres to straight from Lagrange. Second, the facility is never ever topmost. This adheres to by observing that if the facility is not the whole team, allow x be a component not in the facility. After that the centralizer of x is not the whole team, ever since x would certainly remain in the facility. Yet plainly the facility is had in the centralizer of x and also does not have x, so the centralizer of x is a subgroup effectively in between the facility and also the whole team. Currently it is very easy to see that the team has to be abelian, and afterwards it is very easy to show that it has to actually be cyclic (it has a component of order 2, which is less complex to show than Cauchy, and afterwards it simply requires to be revealed that it can not have 2 elemens of order 2, ever since it would certainly have a subgroup of order 4 negating Lagrange). (As a side note, Cauchy is theory for abelian teams is rather straightforward to show simply from expertise of the order of an item of subgroups, and also actually the basic theory after that adheres to from the class formula)

2019-12-03 01:33:55