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

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?

• Do you know the isomorphism theorems? Commented Mar 1, 2022 at 23:12
• @Clayton Yes, some of them. Commented Mar 1, 2022 at 23:15
• Are you allowed Cauchy's Theorem?
– Shaun
Commented Mar 1, 2022 at 23:31
• 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. Commented Mar 1, 2022 at 23:48
• 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. Commented Mar 2, 2022 at 10:21

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 $$C_n$$.

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.

• Right, I removed the comment Commented Mar 4, 2022 at 15:19

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.

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