Show that every group $G$ of order 175 is abelian and list all isomorphism types of these groups. [HINT: Look at Sylow $p$-subgroups and use the fact that every group of order $p^2$ for a prime number $p$ is abelian.]

What I did was this. $|G| = 175$. Splitting 175 gives us $175 = 25 \cdot 7$. Now we want to calculate the Sylow $p$-groups, i.e we want

$$P= n_7: n_7 \equiv 1 mod 7 \hspace{1.5cm} n_7|25$$ $$Q= n_{25}: n_{25} \equiv 1 mod 25 \hspace{1.5cm} n_{25} | 7$$

After listing all elements that are $\equiv 1 mod 7$ and $1 mod 25$ you see that the only (avaliable) ones are $n_7 = n_{25} = 1$. This tells us that both groups $P,Q$ are normal subgroups of $G$. I think, by definition of a normal subgroup, they are abelian and so this tells us that $G$ is abelian. To list all the isomorphism types, we want the semidirect product (SDP) such that

$$P \rightarrow Aut(Q) = C_7 \rightarrow C_{20}$$

As there are no elements of order 7 in $C_{20}$, the only SDP we have is the trivial SDP, i.e the direct product

$$C_7 \times C_{25} \cong C_{175}$$

We know that $175 = 5^2 \cdot 7$ and so multiplying the powers shows us that there are 2 non-isomorphic groups:

$$C_{25} \times C_7$$ $$C_5 \times C_5 \times C_7$$

My question for this is is my reasoning also correct for things like showing the abelian groups? I saw something which said something about $P \cap Q = I_G$ and they used this but I don't understand what it was.

The next question, assuming that I had to possibilites for my $p$ subgroup, i.e $n_p = 1 or x$, how would I go about answering this question? (I am doing a question like this now and am stuck as I have two Sylow $p$-subgroups).

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    $\begingroup$ The fact that $P,Q$ are both normal tells you that $G$ is a direct product of $P$ and $Q$, not that they are abelian. They are abelian because of the hint given in the problem statement. $\endgroup$ – Brandon Carter Jan 9 '13 at 14:09
  • $\begingroup$ From that hint I see that $Q$ is abelian as it has order $25 =5 ^2$, but why is $P$ abelian? $\endgroup$ – Kaish Jan 9 '13 at 14:10
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    $\begingroup$ Because $P$ has order 7, hence is cyclic, and all cyclic groups are abelian. $\endgroup$ – Brandon Carter Jan 9 '13 at 14:11
  • $\begingroup$ If $P$ had order 8, would it be cyclic? EDIT: Don't worry. It might be and it might not be. All groups of prime order are cyclic. I just found that out. $\endgroup$ – Kaish Jan 9 '13 at 14:13

This line seems especially mistaken: "I think, by definition of a normal subgroup, they are abelian and so this tells us that G is abelian." Certainly normal subgroups need not be abelian: for an example you can take the alternating subgroup of the symmetric group for any $n>5$.

The Sylow theorems tell you that $n_7\in \{1,5,25\}$ and that it is 1 mod 7, and so the only possibility is that it is 1.

The Sylow theorems tell you that $n_5\in \{1,7\}$ and that it is 1 mod 5, and so the only possibility is that it is 1.

Thus for both 5 and 7 you have unique (=normal for Sylow subgroups) subgroups. Let's call them $F$ and $S$ respectively. Clearly $FS$ is a subgroup of $G$ of size 175 by the reasoning you gave. (The reason that $F\cap S$ is trivial is that the intersection is a subgroup of both $F$ and $S$, so it must have order dividing both the order of $F$ and of $S$, but the greatest common divisor is 1.)

$S$ is obviously abelian, as it is cyclic (of prime order!). The question is whether or not a group of size 25 must be abelian. There are a lot of ways to see that, but the one that comes to my mind is to say that it definitely has a nontrivial center. If its center $C$ were of order $5$, then $F/C$ would be cyclic of order 5. However, by a lemma (If $G/Z$ is cyclic for a central subgroup $Z$, then $G$ is abelian) $F$ would have to be abelian.

So $G$ is a product of two abelian subgroups, and so is abelian itself.

And also, your conclusion about the two types of abelian groups of order 175 is correct. Initially you wrote that there were "two isomorphic types," but (I edited that to correct it and ) I hope that was just a slip and that you really did mean "two non-isomorphic types".

  • $\begingroup$ I'll be honest and say that it wasn't a slip. I did mean they were isomrphic. Why are they not? I thought that if I'm classifying the isomoprhic types and these groups would be isomorphic. $\endgroup$ – Kaish Jan 9 '13 at 14:26
  • $\begingroup$ @Kaish That's OK, I'm glad it was a real mistake, because now it's a learning opportunity! We want a collection of non-isomorphic groups to be representatives. No other groups matter because they are isomorphic to one of the representatives. Of course, we don't want two isomorphic representatives, because one of them is a "duplicate" of the other! Anyhow, you can see that $C_{25}$ is a cylic group and has several elements of order 25. But $C_5\times C_5$ is not cyclic and it has no elements of order 25! $\endgroup$ – rschwieb Jan 9 '13 at 14:31
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    $\begingroup$ I think by "isomorphic type" Kaish means "isomorphism type", which would be fine. $\endgroup$ – Alex B. Jan 9 '13 at 14:32
  • $\begingroup$ @AlexB. I think you are not seeing the mistake I'm talking about: check the edit history. $\endgroup$ – rschwieb Jan 9 '13 at 14:33
  • $\begingroup$ Found it. $ $ $ $ $ $ $\endgroup$ – Alex B. Jan 9 '13 at 14:36

What you did looks fine, albeit slightly messy and overkill: if you already know there's one unique Sylow $\,5-$subgroup $\,P\,$ of order $\,25\,$ and one single Sylow $\,7-$ subgroup $\,Q\,$ of order $\,7\,$ , both of them abelian, and then you already know:

(1) $\,P,Q\triangleleft G\,$

(2) $\,P\cap Q=1\,$

Then we get

$$|PQ|=\frac{|P||Q|}{|P\cap Q|}=|P||Q|=25\cdot 7=175\Longrightarrow G=PQ\cong P\times Q$$

And since direct product of abelian groups is abelian we're done.

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    $\begingroup$ The OPs argument is not fine, see the comments and the other answer. $\endgroup$ – Alex B. Jan 9 '13 at 14:17
  • $\begingroup$ Yes, that "normal then abelian" thing is a big blunder, but the general thinking is fine, though pretty cumbersome: no need to go to semidirect product to this stuff. Perhaps the OP should also be told that dealing with SDP without knowing these very basic things is dangerous. $\endgroup$ – DonAntonio Jan 9 '13 at 14:21
  • $\begingroup$ @kaish The "messy and overkill" conclusion is also fair, and we don't mean to sound nasty about it. Sometimes we all do too much work :) Just remember to think simple! $\endgroup$ – rschwieb Jan 9 '13 at 14:25
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    $\begingroup$ It can come either as a corollory from the 2nd Isomorphism theorem, although this requires one subgroup to be normal to $G$ - or it can be proven to hold generally for all subgroups of $G$. Probably the most "fundamental" (in that it doesn't assume anything) proof of it that I've seen is on page 14 of "A Course in the Theory of Groups" by Robinson. $\endgroup$ – Andrew D Jan 9 '13 at 14:39
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    $\begingroup$ @Kaish, you can try to prove that formula you ask about as follows: let $\,G\,$ be a group, $\,H,K\leq G\,$ (not necessarily normal!), and denote by $\,H\backslash G\,$ the set of left cosets of $\,H\,$ in $\,G\,$ (this is the quotient group $\,G/H\,$ when $\,H\triangleleft G\,$ ) . Define in an "obvious" way a sets map $\,HK\to H\backslash G\times K\backslash G\,$ , and prove that an element in $\,HK\,$ gets copied to $\,|H\cap K|\,$ elements in the cartesian product $\,H\backslash G\times K\backslash G\,$ ... $\endgroup$ – DonAntonio Jan 9 '13 at 15:01

The fact that $P, Q $ are both normal tells you that $G$ is a direct product of $P$ and $Q$. $P$ is abelian because of the hint given in the problem statement: $|P| = 25 = 5^2$, and $5$ is prime.$|Q| = 7$ with $7$ prime. All groups of prime order are cyclic, and all cyclic groups are abelian, $Q$ is therefore abelian.

Hence $G = P\times Q$, as the direct product of two abelian groups, is therefore abelian.

Just to confirm/answer your first question (listing all possible abelian groups of order 175):

Yes, you correctly showed that if a group $G$ of order 175 is abelian, then it is isomorphic to exactly one of two (non-isomorphic) groups:

$$G \cong C_{175} \cong C_{25} \times C_7$$ $\quad\quad\quad\quad\text{**OR**}$ $$G\cong C_5 \times C_5 \times C_7$$

by the Fundamental Theorem of Finitely Generated Abelian Groups.

Note (added given comment/question below):

$C_{mn} \cong C_m\times C_n$ if and only if $\gcd(m, n) = 1$.

So we have that $C_{175} \cong C_{25}\times C_7$ since $\gcd(25, 7) = 1$.

On the other hand, $C_{25} \times C_7 \ncong (C_5 \times C_5) \times C_7$ since $\gcd(5, 5) \neq 1$

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    $\begingroup$ OHHHHH, so it is isomorhic, its just the groups it is isomorphic to are non-isomorphic? How can you tell this? $\endgroup$ – Kaish Jan 9 '13 at 14:31
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    $\begingroup$ Looks like amWhy hit the nail on the head for you! The fast way is that the first one contains elements of order 25, while the second one doesn't have any elements of order 25. (Isomorphisms preserve orders of elements, of course.) $\endgroup$ – rschwieb Jan 9 '13 at 14:36
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    $\begingroup$ Kaish - G is isomorphic to one or the other, not both. The two possibilities are non-isomorphic, e.g. $C_{25} \ncong C_5\times C_5$ because $\gcd(5, 5) = 5\neq 1$ $\endgroup$ – Namaste Jan 9 '13 at 14:36
  • $\begingroup$ @amWhy: Another very nice answer +1 $\endgroup$ – Amzoti May 8 '13 at 2:38

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