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I have to classify all groups of order 87 and 121. How can I do that? I saw other posts but there is no unified approach that helped me so far...

matt_hematics
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  • $87=3\times 29$ and you should be able to find lots of examples about groups with order $pq$ for two distinct primes. And of course $121=11^2$. – lulu Jul 13 '20 at 14:00
  • See https://math.stackexchange.com/questions/1502186/structure-of-groups-of-order-pq-where-p-q-are-distinct-primes and https://math.stackexchange.com/questions/2546393/classification-of-groups-of-order-p2 – lhf Jul 13 '20 at 14:00
  • Does "$3|(29-1)$" hold? If so, then there would be two groups of order $87$, with one of them being a nontrivial semidirect product $\mathbb{Z}{29} \rtimes \mathbb{Z}{3}$. Otherwise, there would be only one group of order $87$. – Geoffrey Trang Jul 13 '20 at 14:33

2 Answers2

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$\gcd(87, \varphi(87)) = 1$ so there is only one group of order 87: $C_{87}$.

$121 = 11^2$ so there is exactly two groups of this order, $C_{121}$ and $C_{11} \times C_{11}$.

Lemma's used:

Lemma 1: $\gcd(n,\varphi(n)) = 1$ iff there is a unique group of order $n$: $C_n$.

Lemma 2: $n = p^2$ implies there are two groups of order $n$: $C_{p^2}$ and $C_p \times C_p$.

Martin Sleziak
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Let $G$ be a group of order $87=3\times 29$. Using Sylow theorems it follows that $n_3=n_{29}=1$. Since if a $p$-Sylow subgroup is unique it is normal we get that there there are normal subgroups $H,N\trianglelefteq G$ such that $|H|=3$ and $|N|=29$. It it a standard result that if $H, N$ are normal subgroups of $G$ such that $HN=G$, $H\cap N=\{e\}$ then $G\cong H\times N$. So in our case we get: ($H,N$ indeed satisfy what we need)

$G\cong H\times N\cong \mathbb{Z_3}\times\mathbb{Z_{29}}\cong \mathbb{Z_{87}}$

Where the last isomorphism follows from the Chinese remainder theorem. So there is exactly one group of order $87$ up to isomorphism.

Now try to solve for $121=11^2$. A hint: if $p$ is a prime then every group of order $p^2$ is Abelian. Try to use this to show that the groups of order $p^2$ up to isomorphism are $\mathbb{Z_{p^2}}$ and $\mathbb{Z_p}\times\mathbb{Z_p}$.

Mark
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