# Subgroup(s) of a group of order 25

I am working on a problem (self-study) from Artin - 2.8.8 which goes:

"Let G be a group of order 25. Prove that G has at least one subgroup of order 5, and that if it contains only one subgroup of order 5, then it is a cyclic group."

I can see that there is an element of order 5, and this can generate a cyclic subgroup of order 5.

-- So my first question is: what would make one think that there might be more than one subgroup of order 5, and what would they look like. And what would be the criteria for there being only one rather than more than one?

For the second part, I am familiar with a proof that a group of order $p^2$ is abelian, by showing the center is all of G.

-- My second question is how to show G is cyclic - and how does this use the stipulation that there is only one subgroup of order 5 (in the text there is a statement that if there is only one subgroup of a particular order then it is normal). And how does there being more than one subgroup of order 5 prevent G from being cyclic.

Thanks.

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Hint : Suppose $G$ is not cyclic. Then consider the family of subgroups generated by one element. –  user10676 Jan 6 '12 at 18:26
@user10676Forgive my denseness, but maybe you would please elaborate for me. Thanks –  Andrew Jan 6 '12 at 21:32
Suppose $G$ is not cyclic. Take $g \neq 1_G$. Then $<g>$ is of order five (see Alex's explaination). Now take $h \in G-<g>$. Then for the same reason $<h>$ is of order $5$. But by the definition of $h$ we have $<g> \neq <h>$. So $G$ has two subgroups of order $5$. –  user10676 Jan 7 '12 at 14:02
@user10676Thanks, very helpful in itself and regarding Alex's answer. –  Andrew Jan 7 '12 at 22:31
It's a good exercise to prove that for $p$ a prime there are exactly two isomorphism classes of groups of order $p^2$: $\mathbf Z/p^2$ and $\mathbf Z/p \times \mathbf Z/p$. –  Dylan Moreland Jan 8 '12 at 2:43

What might make one think that there may be more than one subgroup of order 5?

Experience: it doesn't take long, as you examine small groups, before you run into groups that have multiple subgroups of the same size: $S_3$, the nonabelian group of order $6$ that corresponds to the bijections of $\{1,2,3\}$ with itself (under composition) has three different subgroups of order $2$. The Klein $4$-group, which is $C_2\times C_2$ (direct product of two cyclic groups of order $2$) has three subgroups of order $2$. The dihedral group with $8$ elements, $D_8$, has five subgroups of order $2$. $C_3\times C_3$, the direct product of two cyclic groups of order $3$, has four subgroups of order $3$; etc. In fact, it's pretty rare for a subgroup $H$ of a group $G$ to be the only subgroup of size $|H|$, except when $|H|=1$ or $|H|=|G|$. It does happen, of course: it happens with cyclic groups. In the alternating group $A_4$ (which has 12 elements), there is a unique subgroup of order $4$. In $S_3$, there is a unique subgroup of order $3$. And so on.

What are the criteria for there being only one [subgroup of a given order] rather than more than one?

Well, of course there's the obvious: $H$ is the only subgroup of $G$ of order $|H|$ if and only if for every subgroup $K$ of $G$, either $K=H$ or $|H|\neq |K|$.

There are some sufficient conditions: if $G$ is finite and cyclic, then $G$ contains a unique subgroup of order $d$ for each $d$ that divides $|G|$. (You may not be aware of this result yet; you'll get to it pretty soon).

But in general it is pretty hard to give a criteria. Essentially, you either just show that the given subgroup is the only one with that order (usually in some relatively ad hoc manner), or you exhibit two distinct subgroups of the same order.

Now, on to the proof. First we show that $G$ must have at least one subgroup of order $5$, no matter what.

Let $g\in G$, $g\neq e$; since $|G|=25$, we can certainly pick an element that is not the identity. By Lagrange's Theorem, the order of $g$ must divide $25=5^2$, so the order of $g$ is either $1$, $5$, or $25$. It cannot be $1$, because the only element of order $1$ is the identity, and we explicitly excluded the identity from our choice of $g$.

If the order of $g$ is $5$, then let $H=\langle g\rangle$. The subgroup generated by $g$ contains exactly as many elements as the order of $g$, so $|H|=\mathrm{order}(g)=5$; and of course, "the subgroup generated by $g$" is a subgroup of $G$. So if the order of $g$ is $5$, then we are done.

The only other possibility is that the order of $g$ is $25$. That means that $g^n=e$ if and only if $25|n$. I claim that the order of $g^5$ is $5$: first, notice that $(g^5)^5= g^{25} = 1$, so the order of $g^5$ divides $5$ (remember that if $x^n=e$, then the order of $x$ divides $n$). So the order of $g^5$ is either $1$ or $5$; it cannot be $1$, because that would mean that $e = (g^5)^1 = g^5$, which would imply, in turn, that the order of $g$ divides $5$. But we are assuming that the order of $g$ is $25$, so $g^5\neq e$, so the order of $g^5$ is not $1$, so the order of $g^5$ must be $5$. Now taking $H=\langle g^5\rangle$ gives a subgroup of order $5$, just as above.

So if $|G|=25$, then there is a subgroup $H$ of $G$ of order $5$.

(In fact, the above argument can be made, exactly the same way, if $|G|=p^2$ for any prime $p$: any such group must have a subgroup of order $p$).

Now on to the second part: we want to prove that our $G$ has a unique subgroup of order $5$ if and only if $G$ is cyclic. This means that we have two prove two implications: if $G$ is cyclic then it has a unique subgroup of order $5$; and if $G$ has a unique subgroup of order $5$ then $G$ is cyclic.

First, we prove that if $G$ is cyclic then it has a unique subgroup of order $5$: since $G$ is cyclic, there is an $x\in G$ such that $G=\langle x\rangle = \{e,x,x^2,\ldots,x^{24}\}$. If $K$ is any subgroup of $G$, other than the trivial subgroup, then there is a smallest positive integer $k$, $1\leq k \lt 25$, such that $x^k\in K$. I claim that $k$ divides $25$ and that $K=\{e,x^k, x^{2k},\ldots,x^{(q-1)k}\}$, where $25 = qk$.

Indeed, using the division algorithm, we can divide $25$ by $k$ with remainder, $25 = qk + r$, $0\leq r\lt k$. Then $$e = x^{25} = x^{qk+r} = x^{qk}x^r = (x^k)^qx^r.$$ Therefore, $x^r = \left((x^k)^q\right)^{-1}$. Note that $x^k\in K$ and $K$ is a subgroup, so $(x^k)^q\in K$, and hence $x^r\left((x^k)^q\right)^{-1}\in K$. But $k$ is the smallest positive integer such that $x^k\in K$, and $0\leq r\lt k$. Since $x^r\in K$, we conclude that the only way this can happen is if $r=0$; and therefore, $25 = qk$, so $k$ divides $25$. And then $k$, $2k,\ldots, (q-1)k$ are all positive and strictly smaller then $25$, so $x^k$, $x^{2k},\ldots,x^{(q-1)k}$ are all distinct. Since $x^{qk} = x^{25}=e$, we get that $K=\{e,x^k,x^{2k},\ldots,x^{(q-1)k}\}$. In particular, $K$ has order $25/k$.

Now, let $H$ be a subgroup of $G$ of order $5$. If the smallest positive $k$ such that $x^k\in H$, then $H$ has order $25/k = 5$. That means that $k=5$, so $H=\{e,x^5,x^{10},x^{15},x^{20}\}$. Note that we concluded this merely from the assumption that $H$ has order $5$; that means that every subgroup of $G$ of order $5$ is equal to $\{e,x^5,x^{10},x^{15},x^{20}\}$, so $G$ has a unique subgroup of order $5$. This proves one of the two implications.

To finish off, we need to show that if $G$ has a unique subgroup of order $5$, then $G$ is cyclic. Let $x\in G$, $x\neq e$. As before, we know that the order of $x$ is either $5$ or $25$; if it is $25$, we are done, because then $\langle x\rangle = G$ and $G$ is cyclic. So assume that $x$ has order $5$. Then $H=\langle x\rangle$ is the unique subgroup of $G$ of order $5$, by assumption. Now, $G-H$ is not empty (there's 20 things in it); let $y\in G-H$. Again, the order of $y$ must be either $1$, $5$, or $25$: it cannot be $1$, because the only element of order $1$ is the identity, and the identity is in $H$, whereas $y\in G-H$. The order cannot be $5$ either: if the order were $5$, then $K=\langle y\rangle$ would be a subgroup of $G$ order $5$; by our assumption $G$ has a unique subgroup of order $5$, so that would mean that $K=H$, since both have order $5$; but then $y\in \langle y\rangle = K = H$, again contradicting our choice of $y$ as an element of $G-H$.

But that only leaves one possibility: that the order of $y$ is $25$. But if the order of $y$ is $25$, then $\langle y\rangle = G$, so $G$ is cyclic, as claimed.

(Again, the argument above works for any group of order $p^2$, where $p$ is a prime; the only thing we used about $25$ is that the only divisors are $1$, $5$, and $25$).

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In the first full paragraph, D8 has one cyclic subgroup of order 4, but the subgroups of order 4. p-groups always have a pretty lush lattice. A4 has a single subgroup of order 4 (and 4 of order 3, and none of order 6). –  Jack Schmidt Jan 8 '12 at 2:23
@Jack: Oops. Quite so... I always forget about those Klein 4-subgroups for some reason. –  Arturo Magidin Jan 8 '12 at 2:47
@Andrew: Yes, that was a typo. Thanks. –  Arturo Magidin Jan 8 '12 at 22:02

Let $g\in G$ be some element other than the identity $e$. Since $\langle g\rangle$ is a nontrivial subgroup, by Lagrange's theorem it must have order $5$ or $25$. In the first case we are done, otherwise $g^5\neq e$ so $\langle g^5\rangle$ has order $5$ as $g^{25}=e$. If there is only one subgroup of order $5$, then unless the group is cyclic $\langle g\rangle = \langle h\rangle$ (as they all have order $5$) for any $g,h\neq e$, so it is easy to see that any $g\in G$ generates the group hence it is cyclic.

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@AlexBeckerThanks. I can follow you through the end of the third sentence. I'm not sure of your logic in the last part. Could you please say some more. –  Andrew Jan 6 '12 at 21:37
@Andrew Suppose the group is not cyclic, so $\langle g\rangle,\langle h\rangle$ are proper subgroups. If we assume $g,h\neq e$ then they must have order $5$, so if we also assume there's only one such group they are equal. Hence $h$ is some power of $g$. But $h$ can be any element of the group other than $e$, and $g^5 = e$, so $G = \langle g\rangle$ hence it is cyclic. –  Alex Becker Jan 6 '12 at 22:09
It is an easy exercise to show that if $c_d(G)$ denotes the number of cyclic subgroups of $G$ of order $d$ then $\displaystyle \sum_{d\mid |G|}c_d(G)\varphi(d)=|G|$ (just partition your group according to the elements orders). Now, if $G$ had only one subgroup of order $5$, then via the fact that all groups of order $5$ are cyclic we can conclude that $c_5(G)=1$. If then $c_{25}(G)=0$ then we'd see that
$$25=|G|=c_1(G)\varphi(1)+4c_5(G)+\varphi(5)=1+4=5$$
Of course, this is ridiculous. So, if $G$ contains only one subgroup of order $5$ then $c_d(G)\ne0$ and so there exists a cyclic subgroup of $G$ of order $25$ which, obviously, must be $G$ itself.
EDIT: Of course, there is nothing special about $5$. The above argument tells you that for a group of order $p^2$, $p$ prime, that being cyclic is equivalent to having one subgroup of order $p$. You will eventually learn that, up to isomorphism, the only groups of order $p^2$ are the cyclic one and the $2$-dimensional $\mathbb{F}_p$-space.