# How many subgroups of order 8 of an abelian group of order 72? [duplicate]

In Fraleigh, there is a question asking if you can say how many subgroups of order 8 an abelian group $G$ of order 72 has, and also if you can say how many subgroups of order 4 that group has. The answer key says there is only one of order 8, but I don't understand why.

I know that $G$ can be decomposed into a direct product of cyclic groups. But both $Z_8 \times Z_9$ and $Z_2 \times Z_4 \times Z_9$ are of order 72; while the first one has a subgroup $Z_8$, the second one has a subgroup of $Z_2 \times Z_4$, and these two groups are not isomorphic.

What am I missing? Also, why couldn't we say how many groups of order 4 $G$ has?

## marked as duplicate by Matt Samuel abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 30 '17 at 2:42

• For each group you have identified (and there are others) you have identified one subgroup of order $8$. Is there an abelian group of order $72$ with more than one such subgroup? – Mark Bennet Jun 29 '17 at 17:11
• Looks like a duplicate. As I answered the target I won't cast the first (and also the last) vote. – Jyrki Lahtonen Jun 29 '17 at 17:24
• Thanks. Is there a way to demonstrate this without reference to Sylow subgroups? This is fairly early on in the text. The only tool I have is the decomposition theorem for finite abelian groups. – ponchan Jun 29 '17 at 17:37
• @ponchan From next time, you should mention what topics you are familiar with in the question itself. This helps people to answer question according to your level and also helps to distinguish your question from possible duplicates and hence prevent your question from getting closed altogether. – Sahiba Arora Jun 29 '17 at 17:44

Note how you actually have said that both of these groups have one subgroup of order $8$. Also note that these are not the only groups of order $72$ so you can not conclude anything by just using just these two groups.

$$72=2^3\cdot3^2$$

As the group is Abelian, therefore all its subgroups are normal. Hence, there exists a unique Sylow-$2$ (say $H$) and Sylow-$3$ (say $K$) subgroup of order $8$ and $9$ respectively.

Now there are 6 Abelian groups of order $72$ upto isomorphism.

Two of these are $\mathbb Z_4 \times \mathbb Z_{18}$ and $\mathbb Z_2\times \mathbb Z_6 \times \mathbb Z_6$. Note that first one has one subgroup of order $4$ and the second one has no subgroup of order $4$. Therefore, you can not conclude the number of subgroups of order $4$.

Edit: Alternatively, if you are not familiar with Sylow theory then you could just list the $6$ Abelian groups of order $72$ and show that each one has exctly one subgroup of order $8$.

• Thank you, but is it possible to explain this without reference to Sylow subgroups? This problem is fairly early on in the text - not even homomorphisms have been explained yet. – ponchan Jun 29 '17 at 17:35
• @ponchan Do you know Structure theorem of Abelian groups? – Sahiba Arora Jun 29 '17 at 17:37
• Yes, that is the only tool I have right now for this purpose. – ponchan Jun 29 '17 at 17:38
• @ponchan Check my edit. – Sahiba Arora Jun 29 '17 at 17:40
• Do we only know that, say, Z2 X Z6 X Z6 has no subgroup of order 4 (and one of order 8) by listing out all the elements and computing their orders? Is there a quicker way? – ponchan Jun 29 '17 at 17:52

If $G$ is Abelian of order $72=9\times 8$ then $G=H_1\times H_2$, with $|H_1|=8$ and $|H_2|=9$. This is a special case of a decomposition theorem of Abelian groups whose order is a product of two coprime numbers. This means that all the elements of $2$-power order lie in $H_1$ and so $H_1$ is the only subgroup of order $8$.

To prove this, consider the map $\phi:G\to G$ given by $\phi(g)=g^8$. Then prove that $G$ is the internal direct product of $\ker\phi$ and $\text{im}\,\phi$, etc.

See you can use Sylow theory . A subgroup of order 8 is basically a Sylow 2-subgroup. Since $G$ is commutative , every Subgroup is normal, hence the 2-sylow subgroup is normal too! Hence unique.

About any subgroup of order 4, you cannot say anything. This is because if $G \cong \mathbb{Z}_{72}$, it is unique. If $G\cong \mathbb{Z}_{2}\times \mathbb{Z}_{4}\times \mathbb{Z}_{9}$, then there are 4 order 4 subgroups, so, it might not be unique. You have to consider different cases.

• Just three subgroups of order $4$ in your second example. Indeed if $H$ is a finite Abelian $2$-group, the number of index $2$ subgroups in $H$ equals the number of elements of order $2$ in $H$, so is one less than a power of $2$. – Lord Shark the Unknown Jun 29 '17 at 17:39