0
$\begingroup$

Let $H$, $G_1$, $G_2$ be groups such that each $G_i$ $(i=1,2)$ has a subgroup $H_i$ isomorphic to $H$. Prove that the direct product $G_1 \times G_2$ has at least 3 distinct subgroups isomorphic to $H$.

I have zero idea how to approach this. I know that $H_1 \times H_2$ is subgroup isomorphic to $H$, but I don't know how to obtain the other two

EDIT: I now know that $H_1 \times H_2$ is wrong and that two groups that work are {$e_1$}x$H_2$, $H_1$x{e$_2$} are the other two, but I am unsure how to show that they are isomorphic to H since they are direct products. Is it enough to show f(uv)=f(u)f(v)

|cite|improve this question
$\endgroup$
  • $\begingroup$ You need to assume that the group $H$ has more than one element; the statement is incorrect if $H$ is a group of order one. $\endgroup$ – bof Oct 30 '17 at 8:15
  • $\begingroup$ $[e_1}\times G_2$ and $G_1\times\{e_2\}$ are not isomorphic to $H,$ it's $\{e_1\}\times H_2$ and $H_1\times\{e_2\}$ that are isomorphic to $H.$ You also need a third group, and it's not $G_1\times G_2,$ $\endgroup$ – bof Oct 30 '17 at 8:19
1
$\begingroup$

The two you are looking for are just the subgroups $H_1 \times \{e_2\}$ and $\{e_1\} \times H_2$, where $e_1, e_2$ are the identity elements of $G_1$ and $G_2$. If $f$ and $g$ are the isomorphisms between $H_1 \leftrightarrow H$ and $H_2 \leftrightarrow H$, respectively, then the isomorphism between $(H_1 \times \{e_2\} )\leftrightarrow H$ is $(h_1,e_1) \mapsto f(h_1)$ and the isomorphism between $(\{e_1\} \times H_2 ) \leftrightarrow H$ is $(e,h_2) \mapsto g(h_2)$.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ My bad, I wrote $H_1, H_2$ instead of $G_1, G_2$ in the problem. So in your answer should the $H_1, H_2$ be $G_1, G_2$? Also how do I show its an isomorphism? Is it just showing f(uv) = f(u)f(v)? $\endgroup$ – Vinny Chase Oct 30 '17 at 3:31
  • $\begingroup$ Sure, just swap out the group names. You'll have to show that it is a bijection and a homomorphism, so one part is showing that it is one-to-one and another is showing that f(uv) = f(u)f(v). $G_1 \times e_1$ is pretty simple, so don't overthink it -- it's basically a copy of $H$, just with an $e$ tacked on that doesn't do anything. $\endgroup$ – Andrew Tindall Oct 30 '17 at 3:57
  • $\begingroup$ Also, your example of $G_1 \times G_2$ is not quite correct. What you should use is the subgroup of $G_1 \times G_2$ that consists only of elements $(g,g)$; where the first element is identical to the second. $\endgroup$ – Andrew Tindall Oct 30 '17 at 3:58
  • $\begingroup$ Ok thank you! Does it matter that the group mentioned in your last comment could be empty? $\endgroup$ – Vinny Chase Oct 30 '17 at 4:01
  • $\begingroup$ Sorry: the elements themselves are not equal, their images under the isomorphisms are. I.e. they correspond to the same element in $H$. This is the subgroup of $G_1 \times G_2$ that consists only of elements $(f^{-1}h,g^{-1}h)$; where the first element is identical to the second under the two isomorphisms $f: G_1 \leftrightarrow H$ and $g: G_2 \leftrightarrow H$ $\endgroup$ – Andrew Tindall Oct 30 '17 at 4:03
3
$\begingroup$

$\def\zee{\mathbb Z}$ Let's do an example: $G_1 = H_1 = \zee$, $G_2 = H_2 = \zee$. and we use additive notation so the group identity is $0$. The three copies of $ \zee$ are $$\zee \times \{0\} = \{(n, 0) : n \in \zee\},$$ $$\{0\} \times \zee = \{(0, n) : n \in \zee\},$$ and the diagonal subgroup, $$D = \{(n, n) : n \in \zee\}.$$ Check that each of these is a subgroup isomorphic to $\zee$. If you understand this example, you can understand the general case. Important note: $\zee \times \zee$ is not isomorphic to $\zee$, and in the general case neither $G_1 \times G_2$ nor $H_1 \times H_2$ is isomorphic to $H$. In fact, if the groups are finite $H_1 \times H_2$ has order $|H_1 \times H_2| = |H|^2$, so the size is wrong.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ I really don't understand this example. Could you explain it? I don't know how to construct the isomorphism between them. Also if the group $G_1$x$G_2$ isn't isomorphic to H how would I come up with something else that is? $\endgroup$ – Vinny Chase Oct 30 '17 at 3:59
  • 1
    $\begingroup$ Can you think of any reasonable map from the set of all integers to the set of all pairs of integers $(n, n)$ with both coordinates the same? $\endgroup$ – fredgoodman Oct 30 '17 at 16:29
  • 1
    $\begingroup$ $f: \mathbb{Z} ->\mathbb{Z} \times \mathbb{Z}$, $f(n) = (n,n)$? $\endgroup$ – Vinny Chase Oct 30 '17 at 16:51
  • 1
    $\begingroup$ Exactly. Now look at the definition of isomorphism and check whether $f$ is an isomorphism. $\endgroup$ – fredgoodman Oct 30 '17 at 17:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.