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Let $G$ a group of order $143$. Prove that $G$ is abelian.

Note: I can't use Sylow theorem, nor any consequence thereof.

My attempt:

I tried to prove that every group of order $143$ is a cyclic group because it is easy to see that every cyclic group is abelian.

Let $H,K$ cyclic subgroups of $G$. By Lagrange theorem we have that $|H|,|K|\in\{1,11,13,143\}$.

Suppose that $|H|=11$ and $|K|=13$, then $H\cap K= e_G$ where $e_G$ is the identity of $G$.

Moreover, I know as $H$ and $K$ are cyclic then $H$ is isomorphic to $\mathbb{Z}_{11}$ and $K$ is isomorphic to $\mathbb{Z}_{13}$ then $H\times K$ is isomorphic to $\mathbb{Z}_{11} \times \mathbb{Z}_{13}$ and $\mathbb{Z}_{11} \times \mathbb{Z}_{13}$ is isomorphic to $\mathbb{Z}_{143}$

Here I'm stuck. Can someone help me?

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  • $\begingroup$ Do you know the isomorphism theorems? $\endgroup$
    – Clayton
    Mar 1, 2022 at 23:12
  • $\begingroup$ @Clayton Yes, some of them. $\endgroup$
    – rcoder
    Mar 1, 2022 at 23:15
  • $\begingroup$ Are you allowed Cauchy's Theorem? $\endgroup$
    – Shaun
    Mar 1, 2022 at 23:31
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    $\begingroup$ The subgroup of order $13$ is normal, because its index is the smallest prime that divides $|G|$. Moreover, its automorphism group is of order $12$, so the action by conjugation of the subgroup of order $11$ is trivial; that is, the subgroups commute elementwise, and you are done. $\endgroup$ Mar 1, 2022 at 23:48
  • $\begingroup$ Look at the possible orders of G/Z(G). If the order is 1, 11, 13, then G/Z(G) is cyclic. The only problem is when Z(G) = {1}. So your problem reduces to checking that this is not the case. $\endgroup$ Mar 2, 2022 at 10:21

3 Answers 3

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We are using Cauchy's Theorem for the existence of an element $h$ of order $11$ and an element $k$ of order $13$. Let $H=\langle h \rangle$ and $K=\langle k \rangle$. Note that $H \cap K=1$ (use Lagrange's Theorem).

Step 1: $K \lhd G$.

Proof If $K$ is not normal we can find a $g \in G$, with $K^g:=g^{-1}Kg \neq K$. Note that, as $K$, also $K^g$ is a subgroup of order $13$ and by Lagrange $K \cap K^g=1$ (here we use that $K^g \neq K$). Hence $$143=|G| \geq |KK^g|=\frac{|K|\cdot|K^g|}{|K \cap K^g|}=|K|\cdot |K^g|=13 \cdot 13 =169$$ a contradiction. $\square$

Step 2: $K \subseteq Z(G)$.

Proof Since $K$ is normal, $G$ acts by conjugation on $K$ leading to a homomorphic embedding $G/C_G(K) \hookrightarrow \text{Aut}(K)$. But $\text{Aut}(K) \cong C_{12}$. So $|G/C_G(K)|$ divides both $12$ and $143$, hence $G=C_G(K)$, equivalent to $K$ being central in $G$. $\square$

Step 3: $G=HK$ (and $H \lhd G$)

Proof $|HK|=\frac{|H|\cdot |K|}{|H \cap K| }= 11 \cdot 13=143$. It follows that $G=HK$. Of course $H$ normalizes $H$ and $K \subseteq Z(G)$ normalizes $H$, whence $H$ is normal in $G$. $\square$

Basically you are done now, since every pair of elements of $G$ commute, indeed $H$, $K$ are abelian and $K$ is central. But one can go one step further.

Step 4: $G \cong H \times K \cong C_{11} \times C_{13} \cong C_{143}.$

Proof (sketch) Every $g \in G$ can be uniquely written as $g=hk$ with $h \in H, k \in K$. This amounts to a bijective homomorphism $G \rightarrow H \times K$, by sending $g$ to $(h,k)$. $\square$

Remark It is well-known that if $n$ is a natural number, then there is only one group of order $n$ if and only if $\text{gcd}(\varphi(n),n)=1$. Of course this group needs to be isomorphic to $C_n$.

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This one doesn't use Sylow (as desired), nor Cauchy. Just the class equation.

In general, if $|G|=pq$, with $p$ and $q$ distinct primes, then $G$ has center either trivial or the whole group (see here). In your case, $p=11$ and $q=13$, or viceversa.

If $Z(G)$ is trivial, then the class equation reads: $$pq=1+k_pp+k_qq \tag 1$$ where $k_i$ is the number of conjugacy classes of size $i=p,q$. Now, there are exactly $k_qq$ elements of order $p$ (they are the ones in the conjugacy classes of size $q$). Since each subgroup of order $p$ contributes $p-1$ elements of order $p$, and two subgroups of order $p$ intersect trivially, then $k_qq=m(p-1)$ for some positive integer $m$ such that $q\mid m$ (because in your case $q\nmid p-1$). Therefore, $(1)$ yields: $$pq=1+k_pp+m'q(p-1) \tag 2$$ for some positive integer $m'$; but then $q\mid 1+k_pp$, namely $1+k_pp=nq$ for some positive integer $n$, which replaced in $(2)$ yields: $$p=n+m'(p-1) \tag 3$$ In order for $m'$ to be a positive integer, it must be $n=1$ (which in turn implies $m'=1$). So, $1+k_pp=q$: but this is a contradiction, because in your case $p\nmid q-1$. So we are left with $G$ Abelian.

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    $\begingroup$ Right, I removed the comment $\endgroup$ Mar 4, 2022 at 15:19
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Our Goal: Group of order 143 is cyclic. We know,

143= 11×13

Let, G be a group of order 143.

if G has an element of order 143 then G is cyclic.

Assume, G has no element of order 143. By Cauchy Theorem, G has element of order 11 & 13.

Let, P & Q are two subgroups ofG with order 11 & 13. Claim : G has unique subgroup of order 13 proof:

If not, Q1 & Q2 be two distinct subgroup of order 13 . Then order of Q1Q2 is 169( since their intersection must be trivial) which is greater than 143 ( contradiction).

Hence, G has unique subgroup of order 13. Hence, PQ is defined & subgroup of G.

Order of subgroup PQ= 143 Then ,it is obvious G = PQ

Define, f:PQ---> P×Q

f(pq)-->(p,q)

( show f is an isomorphism.)

Then, G is isomorphic to P×Q Again, P & Q are isomorphic to Z11 & Z13 respectively.

Hence, G is isomorphic to Z11×Z13. So, G is cyclic.

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  • $\begingroup$ Can you sketch the proof about f? I guess you need the normality of Q anyhow. $\endgroup$
    – user1007416
    Mar 6, 2022 at 13:02
  • $\begingroup$ Definitely you are right . I should mention Q is normal. Ok I edit it. $\endgroup$
    – Nope
    Mar 6, 2022 at 13:22

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