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I have multiple questions regarding: https://math.stackexchange.com/a/864985/997899

Q: Prove that any group of order 15 is cyclic.

A: Let $G$ be a group such that $|G| = 15$.

  1. Show that the group contains normal subgroups of order $3$ and of order $5$. Let's call them $H$ and $K$ respectively.

  2. Prove the following fact: If $H$ and $K$ are normal, $H \cap K = \{e\}$, and $G = HK = \{hk : h \in H, k \in K\}$, then $G \cong H \times K$.

Hints for #2:

  • Prove that $H \cap K$ is a subgroup of both $H$ and $K$.
  • Show that $HK$ is a subgroup of $G$.

Once you have done these, you are more-or-less finished.

  1. How to prove: $G=HK=\{hk:h\in H,k\in K\}$? It's mentioned we need to prove first $GK$ is subgroup of $G$ but then what? why this proves it's the same as $G$?

Note: Number of elements in $HK$ isn't $15$ for sure, as different multiplications may result in the same element in the set.

  1. After I finish 1 why does this solve the problem? How does this say at all that $G$ is cyclic, according to what law?
Shaun
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Algo
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2 Answers2

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$HK$ is a group whose subgroups are both $H$ and $K$. Thus, its order is divisible by both $3$ and $5$, i.e. by $15$, which means it is $15$ at least! As it is contained in the group $G$ of order $15$, we must have $G=HK$.

How to then finish the proof? As groups $H$ and $K$ are of prime order, they are cyclic. So $H\cong C_3$ and $K\cong C_5$. Thus, $G=HK\cong H\times K\cong C_3\times C_5\cong C_{15}$.

The last isomorphism above (and generally, if $(m,n)=1$ it is known that $C_m\times C_n\cong C_{mn}$) is one of the equivalent formulations of Chinese Remainder Theorem.

  • What ≅ means? still I don't understand why this finishes the proof. – Algo Nov 28 '21 at 18:54
  • $\cong$ in the writing above (and in your question) means "isomorphic to". And, if you've proven $G\cong C_{15}$ then, obviously, a group isomorphic to a cyclic group is cyclic. –  Nov 28 '21 at 18:56
  • Thanks, last 2 problems in HK to prove it's subgroup I need to prove (ab)^-1 is in HK but that's not correct as (ab)^-1 =b^-1 * a^-1 – Algo Nov 28 '21 at 18:58
  • Any group of prime order is cyclic and any cyclic group of order $n$ is isomorphic to $\mathbb{Z_n}$ – Sourav Ghosh Nov 28 '21 at 18:58
  • @Algo You should've pointed that you also needed help about #2. Your question implied that you understood both #1 and #2 but did not understand how this then completes the proof. (TBC) –  Nov 28 '21 at 18:59
  • (Cont'd) If $h\in H, k\in K$, then $h^{-1}k^{-1}hk\in K$ (as $H$ is normal, $h^{-1}k^{-1}h\in K$ already!) and $h^{-1}k^{-1}hk\in H$ (as $K$ is normal, $k^{-1}hk\in H$ already!) This means that $h^{-1}k^{-1}hk\in H\cap K={e}$, i.e. $h^{-1}k^{-1}hk=e$, i.e. $hk=kh$. In other words, every element of $H$ commutes with every element of $K$. Does that help? –  Nov 28 '21 at 19:02
  • Sorry I didn't understand this one at all. I need to prove b^-1 * a^-1 in hk which you didn't show – Algo Nov 28 '21 at 19:09
  • @Algo I hoped you would be able to finish it yourself. If $a\in H, b\in K$, then $(ab)^{-1}=b^{-1}a^{-1}=a^{-1}b^{-1}\in HK$ because $a^{-1}\in H, b^{-1}\in K$ and therefore they commute! –  Nov 28 '21 at 19:10
  • What about xy * zw this can''t be proven to be in HK if x,z in H and the others in K – Algo Nov 28 '21 at 19:14
  • Again, use commutativity. $x, z\in H, y,w\in K$ means $(xy)(zw)=(xz)(yw)\in HK$ as $xz\in H, yw\in K$ and we can swap $y$ with $z$. –  Nov 28 '21 at 19:15
  • And, to preempt your next question, $e=e\cdot e\in HK$ as $e\in H$ and $e\in K$. –  Nov 28 '21 at 19:16
  • ()()=()() that's wrong which commutativity you used? no such thing the commutativity is in H alone and in K alone – Algo Nov 28 '21 at 19:20
  • Please look at my 19 min old comment above. Nobody says that either $H$ is commutative or that $K$ is commutative. But if both are normal subgroups and their intersection is trivial, then every element of $H$ commutes with every element of $K$. In the proof above, we only swapped $y\in K$ with $z\in H$ when we said that $xyzw=xzyw$. –  Nov 28 '21 at 19:22
  • (Aside that $H$ and $K$ in this case are commutative because they are of prime order, hence cyclic. But we don't need to assume that!) –  Nov 28 '21 at 19:24
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Cauchy's theorem (group theory) : Let, $G $ is s a finite group and $p$ is a prime number dividing the order of $G$ (the number of elements in G), then $ G$ contains an element of order $p$.

$G$ be an abelian group . Let $a,b$ be elements of $ G$ with order m and n, respectively. If $m$ and $n$are relatively prime, then show that the order of the element $ab$ is $ mn$.

Hence, $G$ contains an element of order $3$ say $a$ and an element of order $5$ say $b$. Then order of $ab$ is $3×5=15$

Since, $G$ is a group of order $15$ and contains an element $ab$ of order $15$.

Hence, $G=<ab $

Sylow's theorem is a another good options.

But an important corollary, if $order(G) =pq$ ,where$ p, q$ are primes with $p<q$ and $p $ doesn't divides $q-1$ , the the group $G$ is cyclic.

Here, $order(G)=3•5 $ and $3$ doesn't divide $( 5-1)=4$

Sourav Ghosh
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