Groups and Group Actions: Lecture 9

In which we think about homomorphisms, and wonder how many genuinely different groups there are with small orders.

  •  Definitions of a homomorphismisomorphism and automorphism.
  • Proposition 41: Let G, H be groups, let \theta: G \to H be a homomorphism.  Take g \in G, n \in \mathbb{Z}.  Then
    1. \theta(e_G) = e_h;
    2. \theta(g^{-1}) = [\theta(g)]^{-1}; and
    3. \theta(g^n) = [\theta(g)]^n.

    The first followed by considering \theta(e_G \ast_G e_G) in two ways.  The second followed from e_H = \theta(e_G) = \theta(gg^{-1}).  The third falls out using induction and previous parts.

  • Corollary 42: Let G, H be groups, let \theta:G \to H be a homomorphism.  Take g \in G of finite order.  Then o(\theta(g)) divides o(g).  Moreover, if \theta is an isomorphism then o(\theta(g)) = o(g).  We showed that [\theta(g)]^{o(g)} = e_H, and then the first part follows using Lemma 23.  For the second part, we noted that if \theta is injective then [\theta(g)]^k = e_H if and only if g^k = e_G.
  • Lemma 43: Let G be a finite group with even order.  Then G contains an element of order 2.  We defined a relation \sim on G via x \sim y if and only if x = y or x = y^{-1}, and noted that this is an equivalence relation.  The equivalence classes have size 1 (\{g\} for elements that are self-inverse — g^2 = e) or 2 (\{g,g^{-1}\}).  By counting the number of each, and remembering that equivalence classes partition the set, we saw that the number of classes of size 1 is even.  Since it’s also at least 1, there must be a non-identity element that’s self-inverse.
  • Theorem 44: Let p be an odd prime.  Let G be a group with order 2p.  Then G is isomorphic to C_{2p} or D_{2p}.  If G is cyclic, then G is isomorphic to C_{2p}, so we supposed that G contains no element of order 2p.  We showed that G contains an element x of order 2, and another y of order p, and then G = \{e, y, y^2, \dotsc, y^{p-1}, x, xy, xy^2, \dotsc, xy^{p-1}\}.  By considering the order of yx, we found that yx = xy^{-1}, so G is isomorphic to D_{2p}.
  • Definition of a quaternion.

Understanding today’s lecture

It would be good to practise writing out some checks that suitable functions are homomorphisms.  Pick two groups, can you find a homomorphism (an interesting homomorphism!) between them?

If you found the proof of Theorem 44 a bit daunting, then a really good plan would be to work through it in a specific case (e.g. groups of order 6) to see what it says — this is a great way to get insight into a proof.

Can you check that Q_8 really is a group?

Further reading

Wikipedia has a page that lists lots of small groups.

The quaternion group Q_8 can be described in many interesting ways.  Quaternions were first described by Hamilton, who supposedly was so excited by his discovery that he carved the definition on a bridge in Dublin that he happened to be standing on at the time.

You might like to read about the classification of finite simple groupsMarcus du Sautoy touches on this in his popular book Finding Moonshine.

Preparation for Lecture 10

What are the homomorphisms from \mathbb{Z} to itself?

Based on your experience of linear maps between vector spaces, how would you define the kernel and image of a homomorphism?  Can you say anything interesting about them as subsets of their respective groups? (Hint: what can you say about the kernel and image of a linear map?)

We can define a binary operation on (left) cosets of a subgroup H in a group G, by defining (g_1 H) \ast (g_2 H) = (g_1 g_2) H for all g_1, g_2 \in G.  Is this well defined?  That is, if we pick g_1' and g_2' with g_1 H = g_1' H and g_2 H = g_2' H, do we necessarily have (g_1 g_2) H = (g_1' g_2') H?  What conditions do we need to impose on H to make this work?


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: