# Proof that every group of order 30 has a subgroup or order 15 [duplicate]

I'm trying to understand this proof that every group of order $30$ has a subgroup of order $15$:

By Example $37.12$, a group $G$ of order $30$ has a normal subgroup of order $5$ or of order $3$. Suppose that $G$ has a normal subgroup $H$ of order $5$. Then $G/H$ is a group of order $6$, which has a normal subgroup $K$ of order $3$ by Sylow theory. If $λ : G → G/H$ is the canonical homomorphism, then $λ^{−1}[K]$ is a normal subgroup of $G$ of order $3 · 5 = 15$. If $G$ has no normal subgroup of order $5$, then it has a normal subgroup $N$ of order $3$, so $G/N$ has order $10$ and has a normal subgroup $L$ of order $5$. Applying to $L$ the inverse of the canonical homomorphism mapping $G$ onto $G/N$ gives a normal subgroup of $G$ of order $15$

First, why the inverse image doesn't have the same number of elements as $K$ since $\lambda$ is a bijection? Also, why the inverse image has $3.15$ elements?

## marked as duplicate by Dietrich Burde 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 23 '17 at 9:30

• What is Example 37.12?? – David G. Stork Jun 23 '17 at 0:24
• Are just examples that says that there exists subgroups or order 3 and 5 – Guerlando OCs Jun 23 '17 at 0:44
• Please edit your question accordingly. Do not refer to unspecified sources. – David G. Stork Jun 23 '17 at 0:49

If it interests you some how then after you have got that the. Either 3 sylow subgroup or 5 sylow subgroup is normal. Call H to be the 5 sylow subgroup and K to be 3 sylow subgroup. Consider HK. HK is a subgroup of order 15. Note that HK is a subgroup because one of H or K is normal. If both H and K we're just subgroups, then HK is not necessarily a subgroup. For example, Take $S_{3}$ and $H=\{ e, (1 2) \}\$ and $K=\{ e, (2 3) \}\$. HK is not a subgroup in this case.

But as in your solution, I think you can argue as follows

If H is the normal 5-sylow subgroup. Consider the group $G/H$. How does any subgroup look like in this group. It is in one-one to correspondence with the subgroup of G containing H. Any subgroup of $G/H$ therefore looks like $K/H$, where K is a subgroup of G containing H. Now $G/H$ has a subgroup of order 3 call it P. But $P =K/H$ as argued before. So , $|P|=|K|/|H|$. Hence $|K|=|P||H|=15$.

$\lambda$ is not a bijection - it is a homomorphism whose kernel is $H$. Since $|H|=5$, the inverse image of any element of $G/H$ has $5$ elements, so the inverse image of $K$ has $3\cdot 5$ elements.

• Why normality is necessary? – Guerlando OCs Jun 23 '17 at 4:46
• If $H$ is not normal, $G/H$ is not a group! – rogerl Jun 23 '17 at 15:56

If $H$ is a normal subgroup then it is the kernel of a homomorphism.

$\lambda:G\to G/H$

The cannonical map takes G to the cosets of H. Each coset has the same number of elements.

If we take some set $K \in G/H$, and look at the pre-image $|\lambda^{-1}(K)| = |H||K|$

• I didn't understand why the image inverse has that order – Guerlando OCs Jun 23 '17 at 4:28
• Also, why normality is necessary? – Guerlando OCs Jun 23 '17 at 4:45