1. ## Homomorphisms Questions

1) Is (Z14, addition under mod 14) isomorphic to a subgroup of (Z35, addition under mod 35)? Of (Z56, addition under mod 56)?

2) Let f: G --> H
a) Show that if H is abelian and f is one-to-one, then G is abelian.
b) Show that if G is abelian and f is onto, then H is abelian.
c) Show that if f is an isomorphism, then G is abelian iff H is.

3) Let G be the group of nonzero complex numbers under multiplication and let H be the subgroup of GL(2,R) consisting of all matrices of the form:
(a b)
(-b a) (It's one matrix, sorry I don't know how to do that in math formula), where not both a and b are 0. Show that G is isomorphic to H.

2. Hi

An isomorphism is a map which is: 1) bijective, 2) a homomorphism.

Since a bijection between finite sets implies that the sets have the same number of elements, if $\displaystyle H\subseteq \mathbb{Z}_n$ is isomorphic to $\displaystyle \mathbb{Z}_{14},$ then $\displaystyle H$ has order 14 and as a consequence 14 divides $\displaystyle n.$ So that divisibility condition is a necessary condition, and since you're dealing with cyclic groups, try to see why it is also a sufficient condition ( think of what are the subgroups of $\displaystyle \mathbb{Z}_n$ ).

a) and b) are done using the two hypotheses given: for instance, a): $\displaystyle H$ abelian and $\displaystyle f$ one-to-one, let $\displaystyle x,y$ be elements of $\displaystyle G,$ then $\displaystyle \exists a,b\in H, x=f^{-1}(a),\ y=f^{-1}(b),$ so, since $\displaystyle f(xy)=ab$ and $\displaystyle f(yx)=ba,\ xy=f^{-1}(ab)=f^{-1}(ba)=yx$ using injectivity again.

c) can be deduced from a) and b).

Compute $\displaystyle \begin{pmatrix}a&b\\-b&a\\\end{pmatrix}\begin{pmatrix}c&d\\-d&c\\\end{pmatrix}$ . Compute $\displaystyle (a+ib)(c+id)$ ? Any idea to find a isomorphism?

3. Ok, thanks. I'm still kinda confused on #1 and #3 if someone can help.

4. Originally Posted by Janu42
1) Is (Z14, addition under mod 14) isomorphic to a subgroup of (Z35, addition under mod 35)? Of (Z56, addition under mod 56)?
The subgroups of $\displaystyle \mathbb{Z}_n$ are all cyclic groups and it has exactly one subgroup of order $\displaystyle d$ for each $\displaystyle d|n$. Therefore, $\displaystyle \mathbb{Z}_{35}$ has no subgroup of order $\displaystyle 14$ and so $\displaystyle \mathbb{Z}_{14}$ is not isomorphic to a subgroup of $\displaystyle \mathbb{Z}_{35}$. However, $\displaystyle \mathbb{Z}_{56}$ does a subgroup of order $\displaystyle 14$ since $\displaystyle 14|56$ and so $\displaystyle \mathbb{Z}_{14}$ is isomorphic to a subgroup of $\displaystyle \mathbb{Z}_{56}$.

3) Let G be the group of nonzero complex numbers under multiplication and let H be the subgroup of GL(2,R) consisting of all matrices of the form:
(a b)
(-b a) (It's one matrix, sorry I don't know how to do that in math formula), where not both a and b are 0. Show that G is isomorphic to H.
Define the function, $\displaystyle \begin{bmatrix}a&b\\-b&a\end{bmatrix} \mapsto a+bi$.