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Question. Are all groups of order 175 abelian?

I can show that there exists only one Sylow 5-subgroup of order 25, call it $H$, and one Sylow 7-subgroup of order 7, denote $K$.

I know that $K$ is cyclic, and thus abelian. I know that $|H| = p^2$, where $p=5$ is prime, and so $H$ is abelian too. I also know the $|G| = |H| \cdot |K|$.

Further, I know that $G$ happens to be the direct product of these two groups as they intersect trivially, and this completes the proof.

Could somebody please explain:

  1. Why the group is the direct product, is this always so if the groups intersect trivially, and the product of the orders of subgroups matches the group order?

  2. Why the direct product is abelian. Is this always the case if the subgroups $H$ and $K$ intersect trivially, or is it because they are both abelian too?

  3. Anything else I should know?

Thanks

the_fox
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rk101
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  • Ad 2) The direct product of abelian groups is abelian. As the product is direct by (1), $G \cong H \times K$ is abelian. – martini May 21 '12 at 11:07
  • Only because they intersect trivially? – rk101 May 21 '12 at 11:10
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    For (1) , you need normality too. As in, you need $H$ and $K$ to be both normal (i.e. there is precisely one Sylow $p$-group and one Sylow $q$ groups where $|G|=p^aq^b$). This works, as for $G$ to be a direct product you need to find two normal subgroups $H$ and $K$ which intersect trivially (which you have) and such that $G=HK$ (which you also have, but proving this is slightly more subtle - but I am sure this is not outwith your abilities. For (3), there are many things you should know. See, for example, http://www.bbc.co.uk/news/ and http://www.tearfund.org/) – user1729 May 21 '12 at 11:12
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    (Also, for (1), let $G$ be a group where $G=HK$ and $H\cap K=1$. If both of $H$ and $K$ are normal $G$ is called a direct product of $H$ and $K$, $G=H\times K$, if one one is normal ($H$ say) then $G$ is called a semidirect direct product of $H$ and $K$, $H\rtimes K$, while if neither is normal then $G$ is called a Zappa-Szep product of $H$ and $K$, $G=H\bowtie K$ (see my answer here http://math.stackexchange.com/questions/107781/has-this-generalized-semidirect-product-been-studied/107788#107788).) – user1729 May 21 '12 at 11:16
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    @rk101, see also http://math.stackexchange.com/questions/67407/group-of-order-15-is-abelian – Nicky Hekster May 21 '12 at 22:28

2 Answers2

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Your questions:

1) This is not enough: it must be also that both sbgps. are normal in G

2) Because they're abelian, too.

3) Lots more, as anyone else...but not for this particular question, imo.

DonAntonio
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Write $175$ as $5^2 \cdot 7$. Then, we can see that neither $5$ nor $5^2=25$ is congruent to $1$ mod $7$, and $7$ is not congruent to $1$ mod $5$. Hence, by the general characterization of numbers $n$ for which all groups of order $n$ are nilpotent, it is true that all groups of order $175$ are nilpotent. Since $175$ is cubefree (but not squarefree), it follows that all groups of order $175$ are abelian (but not necessarily cyclic).

Geoffrey Trang
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