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I am trying to classify all groups of order 75. I am only allowed to use Sylow's theorems, and other basic facts (such as what groups of order $p^2$, $2p$ are isomorphic to, basic facts about direct products, etc).

My attempt was the following:

We have that $75 = 3 \cdot 5^2$.

Sylow's third theorem tells us that the number $n_5$ of Sylow $5$-subgroups is congruent to $1 \bmod 5$, and must divide $3$. Therefore $n_5 = 1$, so we must have a normal (by Sylow's second theorem) subgroup of order $5^2 = 25$.

Similarly we get that $n_3$, the number of Sylow $3$-subgroups, is either $1$ or $25$. In the case that $n_3 = 1$ it isn't hard to show that that the group is isomorphic to either $\mathbb Z_{25} \times \mathbb Z_3 \simeq \mathbb Z_{75}$, or to $(\mathbb Z_5 \times \mathbb Z_5) \times \mathbb Z_3 \simeq \mathbb Z_5 \times \mathbb Z_{15}$.

In the other case when $n_3 = 25$, we have $25$ subgroups where the pairwise intersection of any two is $\{e\}$. This accounts for $1 + 25 \times (3-1) = 51$ distinct elements (including the identity) of the group, leaving $24$ elements.

This is all I know for this case, and I don't know how to proceed. I appreciate any assistance.

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  • $\begingroup$ When we have $n_3 = 1$ shouldn't we have only the case that $G \cong \mathbb{Z}_3\times\mathbb{Z}_5 \times \mathbb{Z}_5\times$? $\endgroup$ – RGS Jan 28 '18 at 20:29
  • $\begingroup$ @RGS When $n_3=1$, $G$ is isomorphic to the direct product of two groups of order $25$ and $3$. The groups of order $25$ are $\mathbb Z_{25}$ and $\mathbb Z_{5} \times \mathbb Z_5$, and the only group of order $3$ is clearly $\mathbb Z_3$. $\endgroup$ – Luke Collins Jan 28 '18 at 20:35
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There are three different groups of order $75$. Of course we have the $2$ abelian groups $C_5\times C_5\times C_3$ and $C_{25}\times C_3$. For the non-abelian, see

Construct a non-abelian group of order 75

A detailed proof about the full classification, using Sylow, is given here, page $3$, part 5.5.8. It shows in particular, that the non-abelian group constructed in the duplicate is the only one.

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  • $\begingroup$ Thanks for this. Can it be done without automorphisms? This was in a past test paper, our lecturer skipped automorphisms this year, perhaps that's why I wasn't able to do the full classification. $\endgroup$ – Luke Collins Jan 28 '18 at 20:32
  • $\begingroup$ Is that denoting the totient function? We didn't use that in this course, so yeah probably this is something we're not expected to do. Thanks for the help. $\endgroup$ – Luke Collins Jan 28 '18 at 20:37
  • $\begingroup$ I suppose you did have Euler's totient function in the course, when you had the group $(\mathbb{Z}/n\mathbb{Z})^{\times}$. If not, then this is a good opportunity. $\endgroup$ – Dietrich Burde Jan 28 '18 at 20:39
  • $\begingroup$ We had only discussed it in a previous course on groups, in the 'applications' part, we proved Euler's theorem. I'm not sure what you mean by the $^\times$ notation there. $\endgroup$ – Luke Collins Jan 28 '18 at 20:43
  • $\begingroup$ If $R$ is a ring ring, then $R^{\times}$ denotes the group of units, sometimes also denoted by $U(R)$. For $R=\mathbb{Z}/n\mathbb{Z}$ the group of units has $\phi(n)$ elements. Euler's Theorem then says that $a^{|R^{\times}|}=1$. $\endgroup$ – Dietrich Burde Jan 28 '18 at 20:45

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