# Classification of groups of order 30 [duplicate]

How do I find all the groups of order 30? That is I need to find all the groups with cardinality 30. I know Sylow theorems.

• If this is homework, please say so. It will help (us help) you if you show your working so far too :) – Shaun Nov 16 '13 at 14:26
• @Shaun It's not an homework btw. I have found all the sylow subgroups and have also found that either sylow-3 or sylow 5 subgroup is Normal. – Shiva Prakash Nov 16 '13 at 14:28
• Try looking at the subgroup of order 15. – Chris C Nov 16 '13 at 14:29

Let $G$ be a group of order 30 and let $n_p$ be the number of its $p$-Sylow subgroups. By Sylow, $n_3\in\{1,10\}$ and $n_5\in\{1,6\}$. It is not possible that $n_3 = 10$ and in the same time $n_5 = 6$, since otherwise, there would be 20 elements of order 3 and 24 elements of order 5, contradicting the fact that $G$ contains only 30 elements.

So either the $3$-Sylow subgroup $P_3$ or the $5$-Sylow subgroup $P_5$ is normal, implying that $N = P_3 P_5$ is a subgroup of $G$. Because of $\gcd(3,5) = 1$, $P_3\cap P_5 = \{1\}$ and therefore $\lvert N\rvert = 15$. By the classification of $pq$-groups and $5\not\equiv 1\bmod 3$, $N$ is cyclic. Since its index in $G$ is $2$, it is a normal subgroup. Any $2$-Sylow subgroup is a complement of $N$ in $G$.

So $G \cong \mathbb Z/15\mathbb Z \rtimes_{\varphi} \mathbb Z/2\mathbb Z$, where $\varphi : \mathbb Z/2\mathbb Z \to \operatorname{Aut}(\mathbb Z/15\mathbb Z)$ is a group homomorphism. We have $\operatorname{Aut}(\mathbb Z/15\mathbb Z) \cong (\mathbb Z/15\mathbb Z)^\times \cong (\mathbb Z/3\mathbb Z \times \mathbb Z/5\mathbb Z)^\times \cong \mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$. Since $\varphi$ is a homomorphism and the order of $1$ in $\mathbb Z/2\mathbb Z$ is $2$, the order of its image $\varphi(1)$ divides $2$. In $\mathbb Z/2\mathbb Z\times \mathbb Z/4\mathbb Z$ there are four possible images, namely $\varphi(1)\in\{(0,0), (1,0), (0,2), (1,2)\}$. Since $\varphi$ is determined by $\varphi(1)$, this shows that there are at most $4$ isomorphism types of groups of order $30$.

Now look at the following four groups or order $30$: $$\mathbb Z/30\mathbb Z,\quad D_{15},\quad \mathbb Z/5\mathbb Z\times S_3,\quad \mathbb Z/3\mathbb Z\times D_5$$ It is not hard to check that they are pairwise non-isomorphic. So there are exactly these $4$ isomorphism types of groups of order $30$.

You should already have seen that a group of order 15 is cyclic. This follows readily from Sylow's theorems.

You say

I have found all the sylow subgroups and have also found that either sylow-3 or sylow 5 subgroup is Normal.

Suppose there is a normal 5-Sylow subgroup $F$, and let $T$ be a 3-Sylow subgroup. Then $F T$ is a subgroup of order 15, so it is cyclic. Moreover $FT$ has index 2, so it is normal. It follows that $T$ is also normal in $G$. A similar argument applies with 3 and 5 exchanged, so now you know that both a 3- and a 5-Sylow subgroup is normal.

Thus $G$ has a cyclic normal subgroup $FT$ of order 15. The last thing to do is to see how an element $\alpha$ of order 2 can act on $FT$. Consider separately the action of $\alpha$ on $F$ and $T$.

• I think for the conclusion that G has a normal subgroup FT of order 15, there is no need to show that both F and T are normal. The knowledge that one of them is normal, combined with the index 2 argument should be sufficient. – azimut Nov 16 '13 at 15:27
• Could explain the following question, please? How it follows that $T$ is also normal in $G$? – ZFR Mar 7 '18 at 16:10
• $FT$ is cyclic, so $T$ is the unique $3$-Sylow subgroup of $FT$. By Sylow Theorems, all $3$-Sylow subgroups of $G$ are conjugate to $T$. Now $FT$ is normal in $G$, so all $3$-Sylow subgroups of $G$ are in $FT$, so $T$ is the only one, and thus normal. – Andreas Caranti Mar 7 '18 at 19:18

F.N.Cole, J.W.Glover. On groups whose orders are products of three prime factors. American Journal of Mathematics, Vol.15, No.3(1893), pp.191-220.

• Nilpotency comes a while after Sylow's theorems, no? – user1729 Nov 16 '13 at 14:26
• (Also, this cannot be right - I think you mean soluble rather than nilpotent. Your answer implies that there is only one group of order $30$, which is cyclic. This is not the case. For example, $S_3\times C_5$.) – user1729 Nov 16 '13 at 14:31